text:
\section*{Introduction}

Let $R$ be a commutative Noetherian ring and let $\mathrm{Ass}_R(R/I)$ be the set of all prime ideals associated to an ideal $I$ of $R$. An ideal $I \subset R$ is called {\em normally torsion-free} if $\mathrm{Ass}_R(R/I^k) \subseteq \mathrm{Ass}_R(R/I)$  for all $k\geq 1$.
 It is a known fact that if $I$ is a square-free monomial ideal in a polynomial ring $S=K[x_1, \ldots, x_n]$ over a field $K$, then $I$ is normally torsion-free if and only if the ordinary powers of $I$ coincide with the symbolic powers of $I$, for example see \cite[Theorem 1.4.6]{HH1}. Some of the well-known classes of normally torsion-free monomial ideals originate from graph theory.
 Simis, Vasconcelos, and Villarreal showed in \cite{SVV}  that  a finite simple graph is bipartite if and only if its edge ideal is normally torsion-free.  In \cite{GRV}, Gitler, Reyes, and Villarreal showed that the cover ideals of bipartite graphs are normally torsion-free.

Let $G$ be a simple graph with a vertex set $V(G)$. Given a vertex $v \in G$, the closed neighborhood of $v$ is a subset of $V(G)$ consisting of $v$ and all the vertices that are adjacent to $v$. A dominating set of $G$ is a subset of $V(G)$ that intersects  the closed neighborhoods of all $v \in V(G)$. In the 1960s, Berge and Ore mathematically formulated the concept of domination in graphs and it has been  studied widely by many researchers due to its far-reaching and growing applications in various fields including computer sciences, operations research, linear algebra, and optimization. For a given graph $G$, characterizing all the dominating sets of $G$ and finding the smallest size of a dominating set are among the central problems in the domination of graphs. We refer to \cite{HHS} for further concepts related to the domination in graphs. The definitions of closed neighborhood ideals and dominating ideals were introduced in \cite{SM}. The closed neighborhood ideal of $G$, denoted by $NI(G)$, is the square-free monomial ideal generated by the monomials that correspond to the closed neighborhoods of vertices of $G$ and the dominating ideal of $G$, denoted by $DI(G)$, is the square-free monomial ideal generated by the monomials that correspond to the dominating sets of $G$. In \cite{NQBM}, the authors discussed some classes of graphs whose closed neighborhood ideals and dominating ideals are normally torsion-free. In this paper, our motivation is to extend the work carried out in \cite{NQBM} and widely expand the classes of graphs that give normally torsion-free closed neighborhood ideals and dominating ideals.

The breakdown of the content of this paper is as follows: Section~\ref{prem} contains all the required ingredients from commutative algebra, graph theory,  and hypergraph theory. In our work, the main reference for the definitions and notions in hypergraph theory is  Berge's book  \cite{B}. In Section~\ref{section2}, we review the literature in hypergraph theory and discover some interesting classes of Mengerian hypergraphs, and in particular, we focus on the neighborhood hypergraphs of simple graphs. Normally torsion-free square-free monomial ideals  are closely related to Mengerian hypergraphs. Let $\mathcal{H}$ be a simple hypergraph. The edge ideal of $\mathcal{H}$, denoted by $I(\mathcal{H})$, is the ideal generated by the monomials corresponding to the edges of $\mathcal{H}$. A hypergraph $\mathcal{H}$ is called {\em Mengerian} if it satisfies a certain  min-max equation, which is known as the Mengerian property in hypergraph theory or  has the max-flow min-cut property in integer programming. Algebraically, it is equivalent to $I(\mathcal{H})$ being normally torsion-free, as illustrated in \cite[Corollary 10.3.15]{HH1} and \cite[Corollary 3.14]{GVV}.  The balanced hypergraphs and the totally balanced hypergraphs are among the most discussed classes of Mengerian hypergraphs. In simple words, balanced hypergraphs are those hypergraphs that avoid ``special" (hypergraphic) cycle of odd length, and the totally balanced hypergraphs are those hypergraphs that avoid ``special" (hypergraphic) cycle of any length. Their formal definitions are stated in Definition~\ref{hypercycle}. Using the results obtained in \cite{BDS}, as a consequence in Theorem \ref{trampoline}, we establish that if $G$ is a chordal graph with no induced odd incomplete trampoline, then the closed neighborhood ideal and the dominating ideal of $G$ is normally torsion-free.

Totally balanced hypergraphs are of particular interest because they generalize the notion of forests (in the sense of graphs). Moreover, it is shown in \cite[Theorem 2.3]{HHTZ} that totally balanced hypergraphs can be identified as simplicial forests. Simplicial forests  were introduced by Faridi in \cite{Far1} and further explored in \cite{Far2}. In \cite{Fa}, Farber introduced strongly chordal graphs and proved that a graph $G$ is strongly chordal if and only if the neighborhood hypergraph of $G$ is totally balanced. Rephrasing this in the algebraic language, we conclude in Theorem~\ref{stronglychordal} that the closed neighborhood ideals  of strongly chordal graphs are normally torsion-free and sequentially Cohen-Macaulay, and their dominating ideals are normally torsion-free and componentwise linear.

 Next, we examine the closed neighborhood ideals and dominating ideals  of cycles. The closed neighborhood ideals of cycles can be identified as $3$-path ideals (or cubic path ideals as in \cite{AB}) of cycles. To deal with $DI(C_n)$, we first review the case of $t$-path ideals $P_n$, in Section~\ref{tpath}. Here $P_n$ denotes the path on $n$ vertices. Let ${\mathcal H}_t(P_n)$ be the hypergraph whose edges are the paths of length $t-1$ in $P_n$ with $n \geq t$. Then ${\mathcal H}_t(P_n)$ is a simplicial forest as shown in \cite[Theorem 2.7]{JT}  and once again by \cite[Theorem 3.2]{HHTZ}, and we have   ${\mathcal H}_t(P_n)$ is totally balanced. This discussion yields a combinatorial proof of normally torsion-freeness of $t$-path ideals of path graphs. We give algebraic proofs of these statements in  Theorem~\ref{Path.NTF}.

 Section~\ref{DIofcycles} is dedicated to the dominating ideals of cycles. It is shown in \cite{AB} that the 3-path ideal of a cycle $C_n$ is normally torsion-free if and only if $n\in \{3,6,9\}$. The 3-path ideal of $C_n$ is the same as $NI(C_n)$. In the first main result of Section~\ref{DIofcycles}, we prove that

\textbf{Theorem \ref{symbolic-cycles}.} \;
{\em Let $C_n$ be a cycle with $n\geq 3$.  Then $DI(C_n)^t=DI(C_n)^{(t)}$ for all $t \geq 1$ if and only if $n\in\{3,6,9\}$. In particular, $DI(C_n)$ is normally torsion-free if and only if $n\in\{3,6,9\}$. }
\newline

This result shows that $NI(C_n)$ is normally torsion-free if and only if $DI(C_n)$ is  normally torsion-free. The stable sets of associated primes of cover ideals of odd cycles are studied in \cite{NKA}. Inspired by \cite[Proposition 3.6]{NKA}, in Theorem~\ref{NNTF-DI(CN)}, we describe the stable sets  of associated primes of dominating ideals of cycles and prove also that $DI(C_{n})$ have the persistence property. We conclude this section
 with a short argument on the relation between  prime monomial ideals  associated to  $\mathrm{Ass}(DI(G)^s)$ for some $s \geq 1$, and prime monomial ideals associated to  $\mathrm{Ass}(DI(G_\mathfrak{p})^s)$, where  $G_\mathfrak{p}$ stands for the induced graph on $\mathfrak{p}$. Specially, we are interested in knowing  that if   $\mathfrak{p}=(x_{i_1}, \ldots, x_{i_r})\in \mathrm{Ass}(DI(G)^s)$ for some $s\geq 1$, then can we deduce that $G_\mathfrak{p}$ is connected?  We remain it as an open question, see Question \ref{OPEN-QUESTION}.

 \section{Required terminologies}\label{prem}
 \subsection{Algebraic background}
In this section, recall some definitions and notions from commutative algebra. Let  $R$ be  a commutative Noetherian ring and $I$ be   an ideal of $R$. A prime ideal $\mathfrak{p}\subset  R$ is an {\it associated prime} of $I$ if there exists an element $v$ in $R$ such that $\mathfrak{p}=(I:_R v)$, where $(I:_R v)=\{r\in R \mid  rv\in I\}$. The  {\it set of associated primes} of $I$, denoted by  $\mathrm{Ass}_R(R/I)$, is the set of all prime ideals associated to  $I$.
The minimal members of $\mathrm{Ass}_R(R/I)$ are called the {\it minimal} primes of $I$, and $\mathrm{Min}(I)$ denotes the set of minimal prime ideals of $I$. Moreover, the associated primes of $I$ which are not minimal are called the {\it embedded} primes of $I$. If $I$ is a square-free monomial ideal, then $\mathrm{Ass}_R(R/I)=\mathrm{Min}(I)$, for example see \cite[Corollary 1.3.6]{HH1}.   Let $I$  be an ideal  of $R$  and  $\mathfrak{p}_1, \ldots, \mathfrak{p}_r$  be   the minimal primes of $I$. When there is no confusion about the underlying ring, we will denote the set of associated primes of $I$ simply by $\mathrm{Ass}(R/I)$ or $\mathrm{Ass}(I)$.
Given an integer $n \geq 1$, the {\it $n$-th symbolic power} of $I$  is defined to be the ideal
$$I^{(n)} = \mathfrak{q}_1  \cap \cdots \cap  \mathfrak{q}_r,$$
where $\mathfrak{q}_i$  is the primary component of $I^n$  corresponding to $\mathfrak{p}_i$.

The ideal $I \subset R$ is said to have the {\it persistence property} if $\mathrm{Ass}(R/I^k)\subseteq \mathrm{Ass}(R/I^{k+1})$ for all positive integers $k$. Brodmann in \cite{Br} showed that in a Noetherian ring $R$ the set of associated prime ideals $\mathrm{Ass}(R/I^s)$ for the powers of an ideal $I$ stabilizes for $s \gg 0$. In other words, there exists the  minimum integer $k>0$ such that $\mathrm{Ass}(R/I^s)=\mathrm{Ass}(R/I^{s+1})$ for all $s\geq k$. This stable set of associated prime ideals is denoted by $\mathrm{Ass}^{\infty}(I)$.

\begin{definition}{\em
An ideal $I$ is called {\it normally torsion-free} if  $\mathrm{Ass}(R/I^k)\subseteq \mathrm{Ass}(R/I)$, for all $k\geq 1$. If $I$ is a square-free monomial ideal, then $I$ is normally torsion-free if and only if $I^k=I^{(k)}$   for all $k \geq 1$  (cf.  \cite[Theorem 1.4.6]{HH1}).
}
 \end{definition}

\begin{definition}{\em Let $R$ be a unitary commutative ring and $I$ an ideal in $R$. An element $f\in R$ is {\it integral} over $I$, if there exists an equation
 $$f^k+c_1f^{k-1}+\cdots +c_{k-1}f+c_k=0 ~~\mathrm{with} ~~ c_i\in I^i.$$
 The set of elements $\overline{I}$ in $R$ which are integral over $I$ is the
 {\it integral closure} of $I$. The ideal $I$ is {\it integrally closed}, if $I=\overline{I}$, and $I$ is {\it normal} if all powers of $I$ are integrally closed, refer to \cite{HH1} for more information. The notion of integrality for a monomial ideal $I$ can be described in a simpler way as following: a monomial $u \in  R=K[x_1, \ldots, x_n]$ is integral over $I\subset R$ if and only if there exists an integer $k$ such that  $u^k \in I^k$, see \cite[Theorem 1.4.2]{HH1}.
 }
\end{definition}

Let $R=K[x_1, \ldots, x_n]$ be a polynomial ring over a field $K$. An ideal $I\subset R$ is called a monomial ideal if it admits a generating set consisting of monomials. Furthermore, $I$ is said to be a square-free monomial ideal if it is generated by square-free monomials. Throughout the following text, the unique minimal generating set of a monomial ideal $I$ will be denoted by $\mathcal{G}(I)$. The {\em support} of a monomial $u$, denoted by $\mathrm{supp}(u)$, is the set of variables that divide $u$. Moreover, for a monomial ideal $I$, we set $\mathrm{supp}(I)=\bigcup_{u \in \mathcal{G}(I)}\mathrm{supp}(u)$. Given a square-free monomial ideal $I\subset R$,  the {\it Alexander dual}  of  $I$, denoted by $I^\vee$,  is given by
 $$I^\vee= \bigcap_{u\in \mathcal{G}(I)} (x_i~:~ x_i \in \mathrm{supp}(u)).$$

\subsection{Graph theory}
Let $G$ be a finite simple undirected graph with the vertex set $V(G)$ and the edge set $E(G)$. The degree of a vertex $v \in V(G)$ is denoted by $\deg(v)$ and it represents the number of vertices adjacent to $v$. A {\em subgraph} of $G$ is a graph $H$ with $V(H) \subseteq V(G)$ and $E(G)\subseteq E(H)$. A subgraph $H$ of $G$ is called an {\em induced} subgraph if it contains all the edges of $G$ that have both vertices in $H$.

A set $T\subseteq V(G)$ is called a {\em vertex cover} of $G$ if it intersects every edge of $G$ non-trivially. A vertex cover is called minimal if it does not properly contain any other vertex cover of $G$. For each vertex $v \in V(G)$, the {\em closed neighborhood} of $v$ in $G$ is defined as follows:
\[
N_G[v]=\{u \in V(G) : \{u,v\} \in E(G)\} \cup \{v\}.
\]
When there is no confusion about the underlying graph, we will denote $N_G[v]$ simply by $N[v]$. A subset $S\subseteq V(G)$ is called a  {\em dominating set} of $G$ if $S \cap N[v] \neq \emptyset$   for all $v\in V(G)$. A dominating set is called {\it minimal} if it does not properly contain any other dominating set of $G$. A minimum dominating set of $G$ is a minimal dominating set with the smallest size.  The {\em dominating number} of $G$, denoted by $\gamma(G)$, is  the size of its minimum dominating set, that is,
\[
\gamma(G)=\mathrm{min} \{|S|: S \text{ is a minimal dominating set
of } G\}.
\]

The dominating sets and domination numbers of graphs are
 well-studied topics in graph theory.  Further   information on dominations in graphs can be found in \cite{HHS}.

 In general,  square-free monomial ideals  can be associated with graphs in many different ways. We recall some commonly known definitions in this context. Let $V(G)=\{x_1,x_2,\ldots, x_n\}$ and $R=K[x_1, \ldots, x_n]$ be the polynomial ring in $n$ variable over a field $K$. The {\em edge ideal} of $G$, denoted by $I(G)$, is
\[
I(G)=(x_ix_j : \{x_i,x_j\} \in E(G)).
\]
The {\em cover ideal} of $G$, denoted by $J(G)$, is
\[
J(G)=(\prod_{x_i \in T}x_i: T \text{ is a minimal vertex cover of } G).
\]
In \cite{SM}, the {\em closed neighborhood ideal} of $G$ has been introduced as
 \[
NI(G)= (\prod_{x_j \in N[x_i]} x_j:  x_i \in V(G)).
\]
Moreover, in \cite{SM}, the dominating ideal of $G$ is defined as
\[
DI(G)=(\prod_{x_i \in S}x_i: S \text{ is a minimal dominating set of } G).
\]
It is a well-recognized fact that $J(G)$ is the Alexander dual of $I(G)$ (cf.  \cite[Lemma 9.1.4]{HH1}). It is shown in \cite[Lemma 2.2]{SM} that $DI(G)$ is the Alexander dual of $NI(G)$.

A {\em path} in $G=(V(G), E(G))$ is a sequence of distinct vertices $x_{i_1},\ldots, x_{i_{t}}$ such that $\{x_{i_j}, x_{i_{j+1}}\}\in E(G)$ for $j=1,\ldots,t-1$.  The {\em length}  of a path is the number of edges in it. We will refer to a path of length $t-1$ as a $t$-path. If a graph $G$ itself is a path on $n$ vertices, then it is denoted by $P_n$.
A {\em cycle}, denoted by $C_n$, is a graph with $V(C_n)=\{x_{1},\ldots, x_{n}\}$ and
\[
E(C_n)= \{\{x_{i}, x_{i+1}\}:  i=1, \ldots, n-1\}\cup\{\{x_1,x_n\}\}.
\]

A graph $G$ is said to be {\em chordal} if it has no induced cycles of length greater than four. Equivalently, a graph $G$ is chordal if and only if it admits a perfect elimination ordering of its vertices. A {\em perfect elimination ordering} of $G$ is an ordering $v_1, \ldots, v_n$ of $V(G)$ with the property that for each $i,j$, and $l$, if $i<j$, $i<l$ with $v_l, v_j \in N_G[v_i]$, then $v_l \in N_G[v_j]$. Note that  chordal graphs are sometimes referred to as {\em triangulated graphs}.

\subsection{Hypergraph theory} Here, we recall some notions from hypergraph theory. A finite {\it hypergraph} $\mathcal{H}$ on a vertex set $V({\mathcal{H}})=\{x_1,x_2,\ldots,x_n\}$ is a collection of edges  $E({\mathcal{H}})=\{ E_1, \ldots, E_m\}$ with $E_i \subseteq V({\mathcal{H}})$, for all $i=1, \ldots,m$.  The {\em incidence matrix} of $\mathcal{H}$ is a $n \times m$ matrix $A=(a_{ij})$ such that $a_{ij}=1$ if $x_i \in E_j$, and $a_{ij}=0$ otherwise.
A hypergraph $\mathcal{H}$ is called {\it simple}, if $E_i \subseteq  E_j$ implies  $i = j$. Simple hypergraphs are also known as {\em clutters}. Moreover, if $|E_i|=d$, for all $i=1, \ldots, m$, then $\mathcal{H}$ is called a {\em $d$-uniform} hypergraph. A $2$-uniform hypergraph $\mathcal{H}$ is just a finite simple graph.

Let $V({\mathcal H})=\{x_1,x_2,\ldots, x_n\}$ and $R=K[x_1, \ldots, x_n]$ be the polynomial ring in $n$ variable over a field $K$. The {\it edge ideal} of $\mathcal{H}$ is given by
$$I(\mathcal{H}) = (\prod_{x_j\in E_i} x_j : E_i\in  E({\mathcal{H}})).$$

A subset  $W \subseteq V({\mathcal{H}})$  is a {\it vertex cover} or {\em transversal} of $\mathcal{H}$ if $W \cap E_i\neq \emptyset$  for all $i=1, \ldots, m$. A transversal $W$ is {\it minimal} if no proper subset of $W$ is a transversal of  $\mathcal{H}$. The family of minimal transversals of $\mathcal{H}$ constitutes a simple hypergraph on $V({\mathcal{H}})$ called the {\em transversal hypergraph} of $\mathcal{H}$, and denoted by $T_r (\mathcal{H})$. The cover ideal of the hypergraph  $\mathcal{H}$,
denoted by $J(\mathcal{H})$, is given by
\[
J(\mathcal{H})=(\prod_{x_i \in W}x_i: W \text{ is a minimal transversal of } \mathcal{H}).
\]
In combinatorial commutative algebra, the transversals of a hypergraph are referred to as covers, and that is why the ideal $J(\mathcal{H})$ is usually called the cover ideal of $\mathcal{H}$. Moreover, similar to the case of edge ideal of graphs, the cover ideal $J(\mathcal{H})$ is  the Alexander dual of $I(\mathcal{H})$, that is,  $J(\mathcal{H})=I(\mathcal{H})^{\vee}$, consult  \cite[Theorem 6.3.39]{V1}.

Let $G$ be a graph with $V(G)=\{x_1, \ldots, x_n\}$. The {\em neighborhood hypergraph} of $G$, denoted by $\mathcal{N}(G)$, is a hypergraph on the vertex set $V(G)$ with edges $N[x_i]$, for all $i=1, \ldots, n$. Note that the edge ideal of ${\mathcal N}(G)$ is the closed neighborhood ideal $NI(G)$. A reader may find a confusing that we use the word ``closed" with ideal but not with hypergraphs. It is because we want to keep the standard notations given in \cite{B} and the naming of ideal given in \cite{SM}. Following \cite[Theorem 6.3.39]{V1}, we can rephrase \cite[Lemma 2.2]{SM} as follows:

\begin{Proposition} \label{transversal}
Let $G$ be a graph. Then $DI(G)$ is the edge ideal of the transversal hypergraph of $\mathcal{N}(G)$.
\end{Proposition}

The {\em $t$-path hypergraph} of $G$, denoted by $\mathcal{H}_t(G)$, is a hypergraph on the vertex set $V(G)$ whose edges are all distinct $t$-paths in $G$. The edge ideal of $\mathcal{H}_t(G)$ is called the {\em $t$-path ideal} of $G$ and is denoted by $I_t(G)$.  When $t=2$, then $I_2(G)$ is simply the {\em edge ideal}  $I(G)$ of $G$. If there is no $t$-path in $G$, then we set $I_t(G)=0$.

\subsection{Simplicial complexes}

 A {\em simplicial complex} $\Delta$ on a set of vertices $V(\Delta)=\{x_1, \ldots, x_n\}$ is a collection of subsets of $V(\Delta)$, with the property that if $F \in \Delta$,  then all subsets of $F$ are also in $\Delta$, including the empty set. An element of $\Delta$ is called a {\em face} of $\Delta$, and
  the maximal faces of $\Delta$ with respect to inclusion are called {\em facets} of $\Delta$. The \textit{facet ideal} of $\Delta$ is given by
\[
I(\Delta)= (\prod_{x_i \in F} x_{i}: F \text{ is a facet of } \Delta).
\]
The set of all facets of $\Delta$ can be viewed as a simple hypergraph. In this case, the definition of the edge ideal of hypergraph consisting of all the facets of $\Delta$ is  the  same as the facet ideal of $\Delta$.

\section{Neighborhood hypergraphs and their transversal hypergraphs}\label{section2}

The edge ideals and cover ideals of bipartite graphs are known to be normally torsion-free, see \cite{SVV} and \cite{GRV}. The bipartite graphs are characterized as the graphs that avoid any cycle of odd lengths. With the help of {\it Macaulay2} \cite{GS}, one can verify that $NI(C_8)$ and $DI(C_8)$ are not normally torsion-free. This shows that one cannot directly extend the results of \cite{SVV} and \cite{GRV} in the case of closed neighborhood ideals and dominating ideals of bipartite graphs. However, if we consider the neighborhood hypergraph ${\mathcal N}(G)$ of a graph $G$ such that there are no hypergraphic cycles in ${\mathcal N}(G)$, then one obtains a similar statement as in the case of edge ideals and cover ideals of bipartite graphs. To establish this, we recall some definitions and notions from hypergraph theory. For more details about these concepts, we refer the  reader to \cite{B}.

Let ${\mathcal H}$ be a simple hypergraph with $n$ vertices and $m$ edges, and let $M$ be the incidence matrix of ${\mathcal H}$.  A hypergraph $\mathcal{H}$ is called {\em Mengerian} if for all ${\bold c} \in  {\mathbb{Z}^n_{+}}$, we have the following equality
\[
\min\{{\bold a} \; \cdot \;{\bold c} : \; {\bold a} \in {\mathbb{Z}^n_{+}},\; M \; \cdot \;{\bold a} \geq {\bold 1}\}=\max\{{\bold b}  \; \cdot \; {\bold 1}:\; {\bold b} \in  {\mathbb{Z}^m_{+}},\;  M^T \; \cdot \; {\bold b} \leq {\bold c}\},
\]
where $\mathbb{Z}_{+}$ denotes the nonnegative integers.
Algebraically, it is equivalent to $I(\mathcal{H})$ being normally torsion-free; it  can be found in  \cite[Corollary 10.3.15]{HH1} and  \cite[Theorem 14.3.6]{V1}. We next  define some of the well-known classes of Mengerian hypergraphs.

\begin{Definition}\label{hypercycle}
{\em A cycle of length $k \geq 2$ in a hypergraph $\mathcal{H}$ is a sequence
\[\mathcal{C}: x_1, E_1, x_2, E_2, \ldots, x_k, E_k, x_1\] such that

\begin{enumerate}
\item $E_1, \ldots, E_k$ are distinct edges of $\mathcal{H}$,
\item $x_1, \ldots, x_k$ are distinct vertices of $\mathcal{H}$,
\item $x_1,x_k \in E_k$ and $x_i,x_{i+1} \in E_i$ for all $i=1, \ldots, k-1$.
\end{enumerate}

The vertices of $\mathcal{C}$ are $x_1, \ldots, x_k$ and the edges of $\mathcal{C}$ are $E_1, \ldots, E_k$. Following \cite[Section 10.3.4]{HH1}, in the following text, a cycle $\mathcal{C}$ of length $k \geq 3$ is said to be {\em special} if all the edges of ${\mathcal C}$ contain  exactly two vertices of $\mathcal{C}$.  Note that if we reduce to the case of graphs,  then the special cycles of length $k \geq 3$ are just the chordless cycles.
}
\end{Definition}
A hypergraph $\mathcal{H}$ is called {\em balanced} if every cycle of $\mathcal{H}$ of odd length $k \geq 3$ has an edge containing at least three vertices of the cycle. Moreover, $\mathcal{H}$ is called {\em totally balanced} if every cycle of $\mathcal{H}$ of length $k \geq 3$ has an edge containing at least three vertices of the cycle, see \cite[Chapter 5]{B}.  In other words, $\mathcal{H}$ is balanced if it does not have special odd cycles and $\mathcal{H}$ is totally balanced if it does not have any special cycles. It  follows from the definition that every totally balanced hypergraph is also balanced.

\begin{Remark}\label{rem:imp}{\em
A well-known result of Fulkerson, Hoffman, and Oppenheim in \cite{FHO} states that balanced hypergraphs are Mengerian. In particular, totally balanced hypergraphs are also Mengerian. Furthermore, \cite[Theorem 19 on  page 207]{B} states that transversal hypergraphs of simple balanced hypergraphs are also Mengerian.
}
\end{Remark}

It is known that the neighborhood hypergraphs of trees are totally balanced, for example see \cite[Example 3 on page 174]{B}.
We can now combine together  Remark~\ref{rem:imp} and Proposition~\ref{transversal} to obtain the following theorem. In particular, it is
an affirmative answer to   \cite[Question 2.8(i)]{NQBM} and   \cite[Question 2.20(i)]{NQBM}.
\begin{Theorem}\label{NI-TREES}
The closed neighborhood ideals  and dominating ideals of trees are normally torsion-free.
\end{Theorem}

It is natural to ask that if there exists a complete characterization of graphs for which their associated neighborhood hypergraphs are balanced or totally balanced. Such graphs are characterized in \cite{Fa} and \cite{BDS}. To describe these graphs we need to recall some definitions from  \cite{Fa} and \cite{BDS}.

\begin{Definition}{\em
Let $n \geq 3$. An {\em incomplete trampoline} is a chordal graph $G$ on $2n$ vertices whose vertex set can be partitioned into two sets, $W=\{y_1, \ldots, y_n\}$ and $U=\{u_1, \ldots, u_n\}$ such that no vertices of $W$ are adjacent to each other, and for each $i$ and $j$, $w_i$ is adjacent to $u_i$ if and only if $i=j$ or $i \equiv j+1 \;(\mathrm{mod\;}n)$. In addition, an incomplete trampoline is called {\em trampoline} if the induced subgraph on $U$ is a complete graph. If $n$ is odd, then (incomplete) trampoline is called an odd (incomplete) trampoline.}
\end{Definition}

Trampolines on $2n$ vertices are sometimes  also mentioned as {\em $n$-sun graphs}. It is shown in \cite[Theorem 2]{BDS} that for a given graph $G$, we have ${\mathcal N}(G)$ is balanced if and only if $G$ is chordal with no induced odd incomplete trampoline. Once again, we can deduce from Remark~\ref{rem:imp} and Proposition~\ref{transversal} the next theorem.

\begin{Theorem}\label{trampoline}
Let $G$ be a chordal graph with no induced odd incomplete trampoline. Then the neighborhood ideal and the dominating ideal of $G$ is normally torsion-free.
\end{Theorem}

\begin{Remark}\label{nonbalanced}
{\em It is crucial to mention that there are Mengerian neighborhood hypergraphs that are not balanced. To see this,  suppose
 that  $C_6=(V(C_6), E(C_6))$ is  the even  cycle graph  of order $6$ on the vertex set $V(C_6)=\{x_1, \ldots, x_6\}$ and the   edge set
$E(C_6)=\{\{x_i, x_{i+1}\}~:~i=1, \ldots, 5\}\cup \{\{x_6,x_1\}\}.$  We  can view  the $C_9$ in a similar way.
The cycle  graphs $C_6$ and $C_9$ are not chordal, and hence  \cite[Theorem 2]{BDS} tells us  ${\mathcal N}(C_6)$ and ${\mathcal N}(C_9)$ are not balanced. Indeed, the edges of ${\mathcal N}(C_6)$ are
\[
N[x_1]=\{x_1, x_2,x_6\}, N[x_2]=\{x_1, x_2, x_3\}, N[x_3]=\{x_2, x_3, x_4\},
\]
\[N[x_4]=\{x_3, x_4, x_5\},N[x_5]=\{x_4, x_5, x_6\},N[x_6]=\{x_5, x_6, x_1\},
\]
 and the sequence $\mathcal{C}: x_1, N[x_2], x_3, N[x_4], x_5, N[x_6], x_1$ gives a special odd cycle in the hypergraph ${\mathcal N}(C_6)$. One can construct a special odd cycle  in ${\mathcal N}(C_9)$ in a similar fashion. In Section~\ref{DIofcycles}, we will show that not only $NI(C_6)$ and $NI(C_9)$ are normally torsion-free but also  their Alexander duals $DI(C_6)$ and $DI(C_9)$ are  normally torsion-free. This highlights that the converse of Theorem~\ref{trampoline} is not true.
}

\end{Remark}

Now, we turn our attention to  the characterization of graphs for which the associated neighborhood hypergraphs are totally balanced.
To accomplish this, one has to recall the following definitions.

A {\em strong elimination ordering} of $G$ is an ordering $v_1, \ldots, v_n$ of $V(G)$ with the property that for each $i,j,k,$ and $l$, if $i<j$, $k<l$ with $v_k, v_l \in N_G[v_i]$ and $v_k \in N_G[v_j]$, then $v_l \in N_G[v_j]$.  Graphs that admit a strong elimination ordering are called {\em strongly chordal graphs}. The strongly chordal graphs were introduced by Farber in \cite{Fa}. They form a subclass of chordal graphs, because a strong elimination ordering is also a perfect  elimination ordering, by letting $i=k$. Among many other characterizations of strongly chordal graphs, it is shown in \cite{Fa} that a graph $G$ is strongly chordal if and only if it is chordal and does not contain any induced trampoline.
   In \cite[Theorem 3.3]{Fa}, a polynomial procedure to recognize strongly chordal graphs and to construct strong elimination orderings is given. Some of the special subclasses of strongly chordal graphs are trees and interval graphs. It is proved in \cite[Theorem 5.2]{Fa} and re-proved in \cite[Theorem 2]{BDS} that a graph $G$ is strongly chordal if and only if ${\mathcal N}(G)$ is totally balanced. We provide a neat  and straightforward proof of the forward implication in the following

\begin{Theorem}\label{strongly chordal}
Let $G$ be a strongly chordal graph. Then $\mathcal{N}(G)$ is totally balanced.
\end{Theorem}
\begin{proof}
Let $G$ be a strongly chordal graph.  To simplify the notation in the following text, for any $p,q \in V(G)$ we write $p<q$ if $p$ appears before $q$ in the strong elimination ordering of $G$.
Let $$\mathcal{C}: x_1, E_1, x_2, E_2, \ldots, x_k, E_k, x_1,$$  be a cycle in $\mathcal{N}(G)$ with $k \geq 3$.  Let  $v_1, \ldots, v_k \in G$ be such that $E_i=N[v_i]$  for all $i=1, \ldots, k$. We can rearrange the edges in $\mathcal{C}$ such that $v_1<v_i$ for all $i=2, \ldots, k$, and $x_2<x_1$.  Since $x_1, x_2 \in E_1=N[v_1]$ and $x_2 \in E_2=N[v_2]$, it follows that $x_1 \in E_2$. This shows that
$\mathcal{C}$ is not a special cycle, and so $\mathcal{N}(G)$ is totally balanced, as claimed.
\end{proof}

Totally balanced hypergraphs have further nice algebraic implications. It follows from \cite[Theorem 2.3]{HHTZ} that a simple hypergraph ${\mathcal H}$ is a simplicial forest if and only if ${\mathcal H}$ is totally balanced. Simplicial forests were introduced by Faridi in \cite{Far1} in the language of simplicial complexes. The facet ideals of simplicial forests possess many nice properties as noted in \cite{Far1} and \cite{Far2}. For example, Faridi proved in \cite[Corollaries  5.5 and  5.6]{Far2} that if $\Delta$ is a simplicial forest,  then its facet ideal $I(\Delta)$ is sequentially Cohen-Macaulay and $I(\Delta)^{\vee}$ is componentwise linear. More details and related definitions can be found in \cite{Far1} and \cite{Far2}.

We summarize the  discussion above  together with Remark~\ref{rem:imp} and Proposition~\ref{transversal}, in the next theorem.

\begin{Theorem}\label{stronglychordal}
Let $G$ be a strongly chordal graph. Then the following statements hold:
\begin{itemize}
\item[(i)]  NI(G) is normally torsion-free.
\item[(ii)]  NI(G) is sequentially Cohen-Macaulay.
\item[(iii)] DI(G) is normally torsion-free.
\item[(iv)]  DI(G) is componentwise linear.
\end{itemize}
\end{Theorem}

\section{The $t$-path hypergraphs of path graphs}\label{tpath}

Let $P_n$ be a path on $n$ vertices. It follows from \cite[Theorem 2.7]{JT} that ${\mathcal H}_t(P_n)$ is a simplicial forest and using \cite[Theorem 3.2]{HHTZ} it can be seen that ${\mathcal H}_t(P_n)$ is totally balanced. Consequently, $I_t(P_n)$ is normally torsion-free. Furthermore,  we conclude   from Remark~\ref{rem:imp}
  that $I_t(P_n)^{\vee}$ is normally torsion-free. We provide another proof of these facts by using algebraic tools. To do this, we first  introduce a technique to construct new normally torsion-free square-free monomial ideals based on the existing ones. This technique will be implemented in the subsequent result.

\begin{theorem} \label{NTF1}
Let $I$ be a normally torsion-free square-free monomial ideal in a polynomial ring  $R = K[x_1, \ldots, x_n]$ and $v$ be a square-free monomial in $R$. Let  $\mathcal{G}(I)=\{u_1, \ldots, u_m\}$ such that $\mathrm{gcd}(v, u_i)$ divides  $\mathrm{gcd}(v, u_{i+1})$  for all $i=1, \ldots, m-1$.   Then the following statements hold:
\begin{itemize}
\item[(i)] $I+vR$  is normally torsion-free.
\item[(ii)] $I+vR$  is nearly normally torsion-free.
\item[(iii)]  $I+vR$ is normal.
\item[(iv)]  $I+vR$ has the strong persistence property.
\item[(v)]  $I+vR$ has the persistence property.
\item[(vi)]  $I+vR$ has the symbolic strong persistence property.
\end{itemize}
\end{theorem}

\begin{proof}
(i)  To simplify the notation, set  $J:=I+vR$.  If  $J\setminus x_t=\mathfrak{m}\setminus x_t$ for some $1\leq t \leq n$, then  we  can write  $J=x_tL +\mathfrak{m}\setminus x_t$. If $L=R$, then $J=\mathfrak{m}$, and the assertion holds. Assume that   $L\neq R$. Let  $h\in \mathcal{G}(L)$. If $x_\alpha
 \mid h$ for some $\alpha \in \{1, \ldots, n\}\setminus \{t\}$, this gives that  $h\in  \mathfrak{m}\setminus x_t$, and hence
 $L\subseteq \mathfrak{m}\setminus x_t$. We thus get  $J=\mathfrak{m}\setminus x_t$, and the proof is over. Thus, we assume that
    $J\setminus x_t \neq \mathfrak{m}\setminus x_t$ for all  $t=1, \ldots,  n$.  In what follows,  we first show that $v\in \mathfrak{p}\setminus \mathfrak{p}^2$ for any $\mathfrak{p}\in \mathrm{Min}(J)$. Pick  an arbitrary element   $\mathfrak{p}\in \mathrm{Min}(J)$. Due to  $v\in J$ and $J\subseteq \mathfrak{p}$, we have  $v \in \mathfrak{p}$.
   On the contrary, assume that  $v\in \mathfrak{p}^2$. Because $v$ is  a  square-free monomial,  this yields that
    $|\mathrm{supp}(\mathfrak{p}) \cap \mathrm{supp}(v)| \geq 2$.    If $\mathrm{gcd}(v, u_m)=1$, then  $\mathrm{gcd}(v, u_i)=1$ for all $i=1, \ldots, m-1$. In this case, the claim can be deduced directly from  Theorem   \cite[Theorem 2.5]{SN}. Let     $\mathrm{supp(gcd}(v,u_m))=\{x_{i_1}, \ldots, x_{i_k}\}$  with    $\{{i_1}, \ldots, {i_k}\}\subseteq \{1, \ldots, n\}.$  Here, one may consider the following cases:
\vspace{1mm}

\textbf{Case 1.}  $x_{i_r} \in \mathfrak{p}$ for some $x_{i_r}\in \{x_{i_1}, \ldots, x_{i_k}\}$.  Since  $|\mathrm{supp}(\mathfrak{p}) \cap \mathrm{supp}(v)| \geq 2$, we can take   $x_\lambda\in  \mathrm{supp}(\mathfrak{p}) \cap \mathrm{supp}(v)$  with  $x_\lambda \neq x_{i_r}$. We claim  that  $x_\lambda \notin  \{x_{i_1}, \ldots, x_{i_k}\}$. Suppose, on the contrary, that
$x_\lambda \in \{x_{i_1}, \ldots, x_{i_k}\}$.  Let    $$c:=\min\{j~:~x_{i_r}\mid u_j, \text{~where~} j=1, \ldots, m\},$$  and
 $$d:=\min\{j~:~x_\lambda\mid u_j, \text{~where~} j=1, \ldots, m\}.$$  If   $c\leq d$ (respectively,  $c>d$), then the assumption yields that
   $I \subseteq \mathfrak{p}\setminus \{x_\lambda\}$ (respectively, $I \subseteq \mathfrak{p}\setminus \{x_{i_r}\}$), and by virtue of
$v \in \mathfrak{p}\setminus \{x_\lambda\} $ (respectively, $v \in \mathfrak{p}\setminus \{x_{i_r}\}$), this gives rise to  $J \subseteq \mathfrak{p}\setminus \{x_\lambda\}$  (respectively,  $J \subseteq \mathfrak{p}\setminus \{x_{i_r}\}$), which contradicts the minimality of
 $\mathfrak{p}$.  Consequently, we conclude that $x_\lambda \notin  \{x_{i_1}, \ldots, x_{i_k}\}$, and thus   $x_\lambda \notin \mathrm{supp}(I)$.
    In the light of  $v \in \mathfrak{p}\setminus \{x_\lambda\}$, this leads to   $J \subseteq \mathfrak{p}\setminus \{x_\lambda\}$, a contradiction to the minimality of $\mathfrak{p}$.
\vspace{1mm}

\textbf{Case 2.}    $x_{i_r} \notin  \mathfrak{p}$ for any  $x_{i_r}\in \{x_{i_1}, \ldots, x_{i_k}\}$.   Select two distinct variables  $x_\alpha, x_\beta \in \mathrm{supp}(\mathfrak{p}) \cap \mathrm{supp}(v)$.  Accordingly, we have $x_\alpha, x_\beta \notin \mathrm{supp}(I)$. We thus  get
 $I \subseteq \mathfrak{p}\setminus \{x_\alpha\} $. On account of  $v \in \mathfrak{p}\setminus \{x_\alpha\} $,  one can  conclude that
 $J \subseteq  \mathfrak{p}\setminus \{x_\alpha\} $. This contradicts the  minimality of $\mathfrak{p}$.

Therefore,  we deduce that $v\notin \mathfrak{p}^2$.  In addition, note that    $J\setminus x_t=I\setminus x_t$ for any $x_t \in \mathrm{supp}(v)$.  It follows also from  \cite[Theorem 3.21]{SN}  that   $I\setminus x_t$ is normally torsion-free  for any $x_t \in \mathrm{supp}(v)$, and so   $J\setminus x_t$ is normally torsion-free  for any $x_t \in \mathrm{supp}(v)$.        Fix $s\geq 1$ and  $x_t \in \mathrm{supp}(v)$.    Suppose, on the contrary, that
$\mathfrak{m}\setminus x_t \in \mathrm{Ass}(R/(J\setminus x_t)^s)$. In view of
$\mathrm{Ass}(R/(J\setminus x_t)^s)=\mathrm{Min}(J\setminus x_t)$, we derive that    $\mathfrak{m}\setminus x_t \in\mathrm{Min}(J\setminus x_t)$, and thus  $J\setminus x_t=\mathfrak{m}\setminus x_t,$ which is a contradiction. We therefore have  $\mathfrak{m}\setminus x_t \notin \mathrm{Ass}(R/(J\setminus x_t)^s)$ for all $s$ and $x_t \in \mathrm{supp}(v)$.   It follows now from \cite[Theorem 3.7]{SNQ} that $J$ is  normally torsion-free, and the proof  is complete.  \par
 (ii)-(vi) can be shown similar to the proof of  \cite[Corollary 4.10]{NQKR}.
\end{proof}

As an application of Theorem \ref{NTF1}, we give the following result.

\begin{proposition} \label{Path.NTF}
The $t$-path ideals of path graphs are normally torsion-free, for all $t\geq 0$.
\end{proposition}

\begin{proof}
Let $P_n$ be a path graph with the vertex set  $V(P_n)=\{x_1, \ldots, x_n\}$ and  the edge set
$E(P_n)=\{\{x_i, x_{i+1}\}~:~ i=1, \ldots, n-1\}.$  Then the  $t$-path ideal of $P_n$ is given by
$I_t(P_n)=(x_{i}x_{i+1}\cdots x_{i+t-1}~:~i=1, \ldots, n-t+1).$
The proof is  by induction on $n$. If $n=t$, then $I_t(P_n)=(x_1x_2 \cdots x_{t})$, and there is nothing to show. Let $n>t$ and that the claim has been proven  for any value less than $n$.   Now, put  $I:=(x_{i}x_{i+1}\cdots x_{i+t-1}~:~i=1, \ldots, n-t)$,  $v:=x_{n-t+1}x_{n-t+2}\cdots x_{n-1}x_n$, and
$u_i:=x_{i}x_{i+1}\cdots x_{i+t-1}$ for all $i=1, \ldots, n-t$. It is easily seen that    $\mathrm{gcd}(v,u_i)=1$ for all $i=1, \ldots, n-2t+1$, and
$\gcd(v, u_i)=x_{n-t+1}x_{n-t+2}\cdots x_{t+i-1}$ for all $i=n-2t+2, \ldots, n-t$.  In particular, this implies that   $\mathrm{gcd}(v, u_i)$ divides  $\mathrm{gcd}(v, u_{i+1})$  for all $i=1, \ldots, n-t-1$.   It  follows also from the induction hypothesis that $I$ is normally torsion-free.  Thanks to     $I_t(P_n)=I+vR$, where $R=K[x_1, \ldots, x_n]$, we deduce  immediately  the assertion from  Theorem \ref{NTF1}(i).
 \end{proof}

In what follows, our aim is to  study the  normally torsion-freeness of the Alexander duals of $3$-path ideals  of path graphs.
For this, we will need the following proposition.

\begin{proposition}\label{Intersection}
Suppose that  $I$ and  $J$  are   two  normally torsion-free square-free monomial ideals in a polynomial ring  $R=K[x_1, \ldots, x_n]$ over a field $K$ such that  $\mathrm{supp}(I) \cap \mathrm{supp}(J)=\emptyset$.  Then $I\cap J=IJ$ is normally torsion-free.
\end{proposition}
\begin{proof}
Due to    $\mathrm{supp}(I) \cap \mathrm{supp}(J)=\emptyset$, we  deduce from \cite[Proposition 1.2.1]{HH1}  that  $IJ=I \cap J$.
In particular,  we have
$\mathrm{Min}(I\cap J)=\mathrm{Min}(I) \cup  \mathrm{Min}(J)$. On the other hand, we know from \cite[Theorem 1.4.6]{HH1} that  a square-free monomial ideal $L$ is normally torsion-free if and only if $L^k=L^{(k)}$ for all $k\geq 1$, where $L^{(k)}$ denotes the $k$-th symbolic power of $L$.
Hence, it is enough for us to show that $(I\cap J)^k=(I\cap J)^{(k)}$ for all $k\geq 1$. To  achieve this, fix $k\geq 1$.
Since $I$ and  $J$  are    normally torsion-free,  one can conclude that $I^k=I^{(k)}$ and $J^k=J^{(k)}$. Thus, we get the following equalities
\begin{align*}
(I \cap J)^{(k)}&=\bigcap_{\mathfrak{p}\in \mathrm{Min}(I\cap J)}((I \cap J)^kR_\mathfrak{p}\cap R)\\
&=\bigcap_{\mathfrak{p}\in \mathrm{Min}(I)}(I^kR_\mathfrak{p}\cap R) \cap \bigcap_{\mathfrak{p}\in \mathrm{Min}(J)}(J^kR_\mathfrak{p}\cap R)\\
&=I^{(k)}\cap  J^{(k)} \\
& =I^k \cap J^k\\
&=(I\cap J)^k.
\end{align*}
This shows that $I\cap J=IJ$ is normally torsion-free, as claimed.
\end{proof}

To prove the next propostion, we will use  the following auxiliary result.

\begin{lemma} (\cite[Lemma 2.17]{NQBM}) \label{NQBM}
Let $I$ be a normally torsion-free square-free  monomial ideal in a polynomial ring $R=K[x_{1},\ldots ,x_{n}]$ with $\mathcal{G}(I) \subset R$. Then the ideal $$L:=IS\cap (x_{n}, x_{n+1}, x_{n+2}, \ldots, x_m)\subset  S=R[x_{n+1}, x_{n+2}, \ldots, x_m],$$  is normally torsion-free.
\end{lemma}

\begin{proposition} \label{NTF-Alexander Dual-Path}
The Alexander duals of $3$-path ideals of path graphs are normally torsion-free square-free monomial ideals.
\end{proposition}

\begin{proof}
Let $P_n$ be a path graph with the vertex set $V(P_n)=\{x_1, \ldots, x_n\}$ and the edge  set $E(P_n)=\{\{x_i, x_{i+1}\}~:~ i=1, \ldots, n-1\}$.
 Then  the Alexander dual of the  $3$-path ideal of the path graph $P_n$, denoted by $I_3(P_n)^{\vee}$,   is given by
$I_3(P_n)^{\vee}=\bigcap_{i=1}^{n-2}(x_i, x_{i+1}, x_{i+2}).$
We show the claim by induction on $n$. Since every prime monomial ideal is normally torsion-free, this implies that the assertion is true for the case in which $n=3$. Let $n>3$ and that  the claim has been shown for any value less than $n$. Our strategy is to use  \cite[Theorem 3.7]{SNQ}. For this purpose, set
 $\mathfrak{p}_i:=(x_i, x_{i+1},  x_{i+2})$ for all $i=1, \ldots, n-2$,  and $$v:=\prod_{k=0}^{[\frac{n-1}{3}]}x_{1+3k}.$$
It  is easy  to check   that $|\mathrm{supp}(v) \cap \mathrm{supp}(\mathfrak{p}_i)|=1$ for all $i=1, \ldots, n-2$. This yields that $v\in \mathfrak{p}_i \setminus \mathfrak{p}_i^2$ for all $i=1, \ldots, n-2$. To complete the argument, one has to verify that
$\mathfrak{m}\setminus x_i \notin \mathrm{Ass}(R/(I_3(P_n)^{\vee}\setminus x_i)^s)$ for all $s$ and $x_i \in \mathrm{supp}(v)$, where $\mathfrak{m}=(x_1, \ldots, x_n)$. Fix $s\geq 1$ and $x_i \in \mathrm{supp}(v)$.  One can  check  that $I_3(P_n)^{\vee}\setminus x_i$, after deleting the redundant components,  can be viewed  as $J_1 \cap J_2 \cap \Gamma$, where $J_1=\bigcap_{r=1}^{\ell_1}\mathfrak{q}_r$ (respectively,  $J_2=\bigcap_{t=1}^{{\ell_2}}\mathfrak{q}'_t$) is  the Alexander dual  of a $3$-path ideal of a  path graph such as $P_1$ (respectively, $P_2$)  with $V(P_1)<n$ (respectively, $V(P_2)<n$), $V(P_1) \cap V(P_2)=\emptyset$ and $\Gamma=\bigcap_{c=1}^{\lambda}(x_{\alpha_c}, x_{\beta_c})$ such that  for any $1\leq r \leq \ell_1$,    $1\leq t \leq \ell_2$,  and
$1\leq c\neq d \leq \lambda$, we have $|\mathrm{supp}(\mathfrak{q}_r) \cap  \mathrm{supp}(x_{\alpha_c}, x_{\beta_c})|\leq 1$,
$|\mathrm{supp}(\mathfrak{q}'_t) \cap  \mathrm{supp}(x_{\alpha_c}, x_{\beta_c})|\leq 1$,
 and
$|\mathrm{supp}(x_{\alpha_c}, x_{\beta_c})  \cap \mathrm{supp}(x_{\alpha_d}, x_{\beta_d})|= 1$.  Furthermore, the inductive hypothesis gives  that $J_1$ and $J_2$ are  normally torsion-free, and Proposition \ref{Intersection} implies that $J_1\cap J_2$ is normally torsion-free. It follows now by  repeated use of  Lemma \ref{NQBM} that $J_1 \cap J_2 \cap \Gamma$  is normally torsion-free, and so  $I_3(P_n)^{\vee}\setminus x_i$ is normally torsion-free. This means that $\mathfrak{m}\setminus x_i \notin \mathrm{Ass}(R/(I_3(P_n)^{\vee}\setminus x_i)^s)$, as required. We conclude from \cite[Theorem 3.7]{SNQ} that $I_3(P_n)^{\vee}$ is normally torsion-free.
This completes the  inductive step, and  therefore  the claim has been shown by induction.
\end{proof}

\section{Stable sets of associated primes of dominating ideals of cycles }\label{DIofcycles}

In this section, we will mainly discuss the normally torsion-freeness  of $DI(C_n)$ and the stable sets of associated primes of $DI(C_n)$. In \cite[Theorem 3.9]{AB},  Alilooee  and  Banerjee  showed that the $3$-path ideal of a graph $G$ is normally torsion-free if and only if $G$  is  a path graph $P_k$ for some $k\geq 3$ or $G$ is a cycle $C_{3k}$ when $k=1, 2, 3$. Observe that the $3$-path ideal of a cycle $C_n$ with $n \geq 3$ coincides with $NI(C_n)$. With this observation, we can rephrase \cite[Theorem 3.9]{AB} in the following way.

\begin{proposition}\label{NI-CYCLE}
Let $C_n$ be a cycle graph with $n\geq 3$. Then  $NI(C_{n})$ is normally torsion-free if and only if $n\in \{3,6,9\}$.
  \end{proposition}

As the first main result of this section, in Theorem~\ref{symbolic-cycles}, we will prove that the  statement above also holds for $DI(C_n)$. To  show   this, we first consider the case of $DI(C_6)$ and $DI(C_9)$ and verify  that they are normally torsion-free.

\begin{lemma}\label{DI(C6+C9)}
The dominating ideals of $C_6$ and $C_9$ are normally torsion-free.
\end{lemma}
\begin{proof}
(i)  Suppose that  $E(C_6)=\{\{x_i, x_{i+1}\} : i=1, \ldots, 5\}\cup \{\{x_6,x_1\}\}.$ Then the dominating ideal of $C_6$ is given by
\begin{align*}
DI(C_6)=& (x_1,x_2,x_3) \cap (x_2,x_3,x_4) \cap (x_3,x_4,x_5) \cap (x_4,x_5,x_6) \cap (x_5,x_6,x_1) \\
& \cap (x_6,x_1,x_2)\\
=&(x_3x_6,x_2x_5,x_1x_4,x_2x_4x_6,x_1x_3x_5).
\end{align*}
We want to demonstrate that   $DI(C_6)$ is normally torsion-free by using  \cite[Theorem 3.7]{SNQ}.  For this purpose, set $v:=x_3x_6$ and $L:=DI(C_6)$.
One can easily see that $v\in \mathfrak{p}\setminus \mathfrak{p}^2$ for any $\mathfrak{p}\in \mathrm{Min}(L)$. Here, we must show
that  $\mathfrak{m}\setminus x_i \notin \mathrm{Ass}(R/(L\setminus x_i)^s)$ for all $s$ and $x_i \in \{x_3, x_6\}.$ Fix $s\geq 1$. First, we establish  $\mathfrak{m}\setminus x_3 \notin \mathrm{Ass}(R/(L\setminus x_3)^s)$, where $\mathfrak{m}=(x_1, \ldots, x_6)$.  According to the  definition of deletion, we obtain
 \begin{align*}
L\setminus x_3=& (x_1,x_2) \cap (x_2,x_4) \cap (x_4,x_5)  \cap (x_5,x_6,x_1)\\
=&(x_2x_5, x_1x_4, x_2x_4x_6).
\end{align*}
 It follows from \cite[Theorem 2.5 and Lemma 3.12]{SN} that $x_4(x_1, x_2x_6)$  is  normally torsion-free. Because  $\mathrm{gcd}(x_2x_5, x_1x_4)=1$ and $\mathrm{gcd}(x_2x_5, x_2x_4x_6)=x_2$, one can derive from Theorem  \ref{NTF1} that $L\setminus x_3$ is normally torsion-free, and so $\mathfrak{m}\setminus x_3=(x_1,x_2,x_4,x_5,x_6) \notin \mathrm{Ass}(R/(L\setminus x_3)^s)$.  Now, we show that
 $\mathfrak{m}\setminus x_6 \notin \mathrm{Ass}(R/(L\setminus x_3)^s)$. A similar computation gives that
 \begin{align*}
L\setminus x_6=& (x_1,x_2) \cap (x_2,x_3, x_4) \cap (x_4,x_5)  \cap (x_5,x_1)\\
=&(x_2x_5, x_1x_4, x_1x_3x_5).
\end{align*}
On account on \cite[Theorem 2.5 and Lemma 3.12]{SN}, we get  $x_1(x_4, x_3x_5)$  is  normally torsion-free. As  $\mathrm{gcd}(x_2x_5, x_1x_4)=1$ and $\mathrm{gcd}(x_2x_5, x_1x_3x_5)=x_5$,  Theorem  \ref{NTF1} implies that $L\setminus x_6$ is normally torsion-free, and hence  $\mathfrak{m}\setminus x_6=(x_1,x_2, x_3, x_4,x_5) \notin \mathrm{Ass}(R/(L\setminus x_6)^s)$. It follows now from \cite[Theorem 3.7]{SNQ}  that $DI(C_6)$ is normally torsion-free, as claimed.

(ii)  Let  $E(C_9)=\{\{x_i, x_{i+1}\} : i=1, \ldots, 8\}\cup \{\{x_9,x_1\}\}.$ Then the dominating ideal of $C_9$ is given by
\begin{align*}
DI(C_9)=& (x_1,x_2,x_3) \cap (x_2,x_3,x_4) \cap (x_3,x_4,x_5) \cap (x_4,x_5,x_6) \cap (x_5,x_6,x_7) \\
& \cap (x_6,x_7,x_8)\cap (x_7,x_8,x_9)\cap (x_8,x_9,x_1) \cap (x_9,x_1,x_2).
\end{align*}
Our strategy is to use \cite[Theorem 3.7]{SNQ}. To do this, put $v:=x_3x_6x_9$ and $L:=DI(C_9)$. It is routine to investigate that  $v\in \mathfrak{p}\setminus \mathfrak{p}^2$ for any $\mathfrak{p}\in \mathrm{Min}(L)$.
  To complete the argument, one has to verify that  $\mathfrak{m}\setminus x_i \notin \mathrm{Ass}(R/(L\setminus x_i)^s)$ for all $s$ and $x_i \in \{x_3, x_6, x_9\}.$ Fix $s\geq 1$. We first  prove that $\mathfrak{m}\setminus x_3 \notin \mathrm{Ass}(R/(L\setminus x_3)^s)$, where $\mathfrak{m}=(x_1, \ldots, x_9)$.  It follows from the definition of deletion  that
 \begin{align*}
L\setminus x_3=& (x_1,x_2) \cap (x_2,x_4) \cap (x_4,x_5)  \cap (x_5,x_6,x_7)  \cap (x_6,x_7,x_8)\cap (x_7,x_8,x_9)\\
& \cap  (x_8,x_9,x_1).
\end{align*}
Set $\alpha:=x_1x_4x_7$ and $A:=L \setminus x_3$. It is easy to see that $\alpha \in \mathfrak{p}\setminus \mathfrak{p}^2$ for any $\mathfrak{p}\in \mathrm{Min}(A)$. We show that  $Q\setminus x_i \notin \mathrm{Ass}(R/(A\setminus x_i)^s)$ for all $s$ and $x_i \in \{x_1, x_4, x_7\}$, where $Q=\mathfrak{m}\setminus x_3$.  Fix $s\geq 1$. We first prove that   $Q\setminus x_1 \notin \mathrm{Ass}(R/(A\setminus x_1)^s)$.   Direct computation gives that
$$A\setminus x_1= (x_6,x_7,x_8)\cap   (x_5,x_6,x_7)  \cap  (x_8,x_9) \cap  (x_4,x_5)  \cap (x_2).$$

It follows from Lemma \ref{NTF-Alexander Dual-Path} that $(x_6,x_7,x_8)\cap   (x_5,x_6,x_7)$ is normally torsion-free, and by virtue of Lemma \ref{NQBM}, we get $(x_6,x_7,x_8)\cap   (x_5,x_6,x_7) \cap  (x_8,x_9) \cap  (x_4,x_5)$ is normally torsion-free. According to Proposition \ref{Intersection}, one has
$A\setminus x_1$ is normally torsion-free, and so  $Q\setminus x_1 \notin \mathrm{Ass}(R/(A\setminus x_1)^s)$. By a similar argument, one can show that
$$A\setminus x_4=(x_2) \cap (x_5)  \cap  (x_6,x_7,x_8)\cap (x_7,x_8,x_9)   \cap  (x_8,x_9,x_1),$$
and
$$A\setminus x_7=(x_1,x_2) \cap (x_2,x_4) \cap (x_4,x_5)  \cap (x_5,x_6)  \cap (x_6,x_8)\cap (x_8,x_9),$$
are normally torsion-free. Hence,  we get $Q\setminus x_4 \notin \mathrm{Ass}(R/(A\setminus x_4)^s)$ and  $Q\setminus x_7 \notin \mathrm{Ass}(R/(A\setminus x_7)^s)$. We deduce from \cite[Theorem 3.7]{SNQ} that $A=L \setminus x_3$ is normally torsion-free, and so
$\mathfrak{m}\setminus x_3 \notin \mathrm{Ass}(R/(L\setminus x_3)^s)$. In what follows, we give the sketch of the proof of normally torsion-freeness of
$B:=L \setminus x_6$  and $C:=L \setminus x_9$. We can rapidly conclude   the following
 \begin{align*}
L\setminus x_6=& (x_4,x_5) \cap (x_5,x_7) \cap (x_7,x_8)  \cap (x_1, x_2, x_3) \cap (x_2, x_3, x_4) \cap (x_1, x_8, x_9)\\
& \cap  (x_1, x_2, x_9),
\end{align*}
and
 \begin{align*}
L\setminus x_9=& (x_7,x_8) \cap (x_8,x_1) \cap (x_1,x_2)  \cap  (x_2,x_3,x_4) \cap (x_3,x_4,x_5) \cap (x_4,x_5,x_6)\\
& \cap (x_5,x_6,x_7).
\end{align*}
It is routine to check that $\alpha=x_1x_4x_7 \in \mathfrak{p}\setminus \mathfrak{p}^2$ for any $\mathfrak{p}\in \mathrm{Min}(B)$ (respectively,
for any $\mathfrak{p}\in \mathrm{Min}(C)$). Furthermore, we have the following
$$B\setminus x_1= (x_4,x_5) \cap (x_5,x_7) \cap (x_7,x_8)  \cap (x_8, x_9) \cap (x_9,  x_2)  \cap   (x_2, x_3), $$
$$B\setminus x_4= (x_5) \cap  (x_7, x_8)  \cap (x_2, x_3) \cap (x_1, x_8, x_9) \cap (x_1, x_2, x_9), $$
$$B\setminus x_7=  (x_5) \cap (x_8)  \cap (x_1, x_2, x_3) \cap (x_2, x_3, x_4) \cap (x_1, x_2, x_9), $$
$$C\setminus x_1=(x_8) \cap (x_2) \cap  (x_3,x_4,x_5) \cap (x_4,x_5,x_6) \cap (x_5,x_6,x_7), $$
$$C\setminus x_4=(x_7,x_8) \cap (x_8,x_1) \cap (x_1,x_2)  \cap  (x_2,x_3) \cap (x_3,x_5) \cap (x_5,x_6),$$
and
$$C\setminus x_7=(x_8) \cap (x_1,x_2) \cap (x_5, x_6) \cap (x_2,x_3,x_4) \cap (x_3,x_4,x_5).$$
 By using  Lemma \ref{NTF-Alexander Dual-Path}, \cite[Theorem 2.5]{SN}, and Lemma \ref{NQBM}, we can conclude that
 $B\setminus x_1$, $B\setminus x_4$, and  $B\setminus x_7$ (respectively,  $C\setminus x_1$, $C\setminus x_4$, and  $C\setminus x_7$)  are normally torsion-free.    This gives rise to  $\mathfrak{m}\setminus x_6 \notin \mathrm{Ass}(R/(I\setminus x_6)^s)$ (respectively,  $\mathfrak{m}\setminus x_9 \notin \mathrm{Ass}(R/(I\setminus x_9)^s)$. Finally,  \cite[Theorem 3.7]{SNQ} gives that  $DI(C_9)$ is normally torsion-free.
\end{proof}

We are now ready to establish  the first main result of this section in the following theorem.

\begin{theorem}\label{symbolic-cycles}
Let $C_n=(V(C_n), E(C_n))$ be a cycle graph of order $n\geq 3$.  Then $DI(C_n)^t=DI(C_n)^{(t)}$ for all $t \geq 1$ if and only if $n\in\{3,6,9\}$. In particular, $DI(C_n)$ is normally torsion-free if and only if $n\in\{3,6,9\}$.
\end{theorem}

\begin{proof}
We first assume that $n\in\{3,6,9\}$.
Let $V(C_n)=\{x_1, \ldots, x_n\}$ and $E(C_n)=\{\{x_i, x_{i+1}\} \; : i=1, \ldots, n-1\}\cup\{\{x_n,x_1\}\}$.
It is easy to see  that  $DI(C_3)=(x_1, x_2,x_3)$ is normally torsion-free. According to  Lemma~\ref{DI(C6+C9)}, it follows that  $DI(C_6)$ and $DI(C_9)$ are also normally torsion-free. To complete the proof, we need to show that  $DI(C_n)^t \neq DI(C_n)^{(t)}$ for some $t\geq 1$ when $n\notin \{3,6,9\}$.  To do this, we divide the proof into three cases. We will repeatedly use the following facts in our proof.
\begin{itemize}
\item[(i)] The domination number of  $C_{3k}$ is $k$, and  the domination number of $C_{3k+1}$ and $C_{3k+2}$ is $k+1$, see  \cite{HHRC} for more information.
 \item[(ii)]  The associated primes of $DI(C_n)$ correspond to the closed neighborhoods of the vertices of $C_n$.
\end{itemize}
Now,  we proceed with the proof.
\vspace{1mm}

\textbf{Case 1.} Let $n=3k$  for some $k \geq 4$. It follows from \cite[Lemma 4]{HHRC} that there are exactly three minimal dominating sets of $C_n$ of cardinality $k$, given by
\[
A=\{x_1, x_4, x_7, \ldots, x_{3k-2}\},  B=\{x_2, x_5, x_8, \ldots, x_{3k-1}\},\; \text{and}
\]
\[  C=\{x_3, x_6, x_9, \ldots, x_{3k}\}.
\]
Our aim is to  construct an element $f$ with $\deg f= 2k+1$ such that none of  $A$ or $B$  or $C$ is in $\mathrm{supp}(f)$. To do this,
set   $h:= x_4x_8\prod_{j=4}^kx_{3j}$ and $g:=\prod_{i=1}^{3k} x_i$.
We claim that $f=g/h \notin  DI(C_{3k})^2$ but $f \in  DI(C_{3k})^{(2)}$. It is routine to check  that $\deg h = k-1$, and hence $\deg f= 3k-k+1= 2k+1$. In addition, from the constructions of $A$, $B$, and $C$, we know that  the only elements of degree $k$  in $\mathcal{G}(DI(C_{3k}))$ are
\[
\{\prod_{x_i\in A}x_i, \prod_{x_i\in B}x_i, \prod_{x_i\in C}x_i\}.
\]
Since $x_4, x_8, x_{12} \notin \mathrm{supp}(f)$, we deduce that   $\prod_{x_i\in A}x_i \nmid f,$ $\prod_{x_i\in B}x_i\nmid f,$ and
$\prod_{x_i\in C}x_i\nmid f$. Therefore, none of the elements of degree $k$ in  $DI(C_{3k})$ divides $f$, and consequently $f \notin DI(C_{3k})^2$.  Here, recall that
$$DI(C_{3k})^{(2)}=\bigcap_{\mathfrak{p}\in \mathrm{Ass}(DI(C_{3k}))}\mathfrak{p}^2.$$
To see  $f \in  DI(C_{3k})^{(2)}$, note that the closed neighborhood set of each vertex of $C_{3k}$ is covered twice by $\mathrm{supp}(f)$.  Indeed, for each $x_i \in C_{3k}$, we have either $x_{i}x_{i+1} \in \mathrm{supp}(f)$, or $x_{i-1}x_{i} \in \mathrm{supp}(f)$, or $x_{i-1}x_{i+1} \in \mathrm{supp}(f)$.
\vspace{1mm}

\textbf{Case 2.}  It follows from  $x_1 x_2 x_3\in DI(C_4)^{(2)}\setminus DI(C_4)^{2}$  that $DI(C_4)^2 \neq DI(C_4)^{(2)}$, and so the assertion is true  for $C_4$. Let $n=3k+1$ for some $k\geq 2$. Let  $f':= x_{3k+1} \prod_{i=0}^{k-1}x_{3i+1}x_{3i+2}.$
Then $\deg f'=2k+1$. By virtue of  the minimal degree of any  generator in $DI(C_{3k+1})^2$ is $2k+2$, this  shows that $f' \notin DI(C_{3k+1})^2$. Moreover, the closed neighborhood set of each vertex of $C_{3k+1}$ is covered twice by $ \mathrm{supp}(f')$, which yields  that $f' \in DI(C_{3k+1})^{(2)}$.
\vspace{1mm}

\textbf{Case  3.}  Let $n=3k+2$ for some $k\geq 1$ and  put  $f'':=\prod_{i=1}^{3k+2} x_i$.
Then $\deg f''=3k+2$. In view of  the minimal degree of any  generator in $DI(C_{3k+2})^3$ is $3k+3$, we get  $f'' \notin DI(C_{3k+2})^3$. Furthermore, the closed neighborhood set  of each vertex of $C_{3k+2}$ is covered three times  by $\supp(f'')$, which gives  that $f'' \in DI(C_{3k+2})^{(3)}$.

Accordingly, the discussion above shows  that if $n\notin \{3,6,9\}$, then   $DI(C_n)^t \neq DI(C_n)^{(t)}$ for some $t\geq 1$. This finishes the  proof.
\end{proof}

In general, it is well-known that the cover ideals of even cycle graphs are always normally torsion-free (cf.  \cite[Corollary 2.6]{GRV}). It has already been shown  in \cite[Theorem 4.5]{NQKR} that the cover ideals of odd cycle graphs are nearly normally torsion-free, see \cite[Definition 1.1]{NQKR}.  Furthermore, the associated primes of powers of cover ideals of odd cycle graphs have already   been  studied in \cite{NKA}, refer to the  proposition below.

\begin{proposition}(\cite[Proposition 3.6]{NKA})
 Suppose that $C_{2n+1}$ is  a cycle graph on the  vertex set $[2n+1]$, $R=K[x_1, \ldots, x_{2n+1}]$ is a  polynomial ring over a field $K$,
 and $\mathfrak{m}$ is the unique homogeneous maximal ideal  of $R$. Then  $$\mathrm{Ass}_R(R/(J(C_{2n+1}))^s)= \mathrm{Ass}_R(R/J(C_{2n+1}))\cup \{\mathfrak{m}\},$$  for all $s\geq 2$. In particular,
 $$\mathrm{Ass}^\infty(J(C_{2n+1}))=\{(x_i, x_{i+1})~: ~ i=1, \ldots 2n\}\cup\{(x_{2n+1}, x_1)\}\cup \{\mathfrak{m}\}.$$
  \end{proposition}

In what follows, our aim is to establish that a similar result as above holds for  the  associated primes of powers of dominating ideals of cycle graphs of any order.
 Before stating the next theorem, we  recall the definition of the  {\it  monomial localization}   of a monomial ideal with respect to a monomial prime ideal.
     Let $I$ be a monomial ideal in a polynomial ring $R=K[x_1, \ldots, x_n]$ over a field $K$. In addition, we  denote by $V^*(I)$ the set of monomial prime ideals containing $I$. Let $\mathfrak{p}=(x_{i_1}, \ldots, x_{i_r})$ be a monomial prime ideal. The {\it monomial localization} of $I$ with respect to $\mathfrak{p}$, denoted by $I(\mathfrak{p})$, is the ideal in the polynomial ring $R(\mathfrak{p})=K[x_{i_1}, \ldots, x_{i_r}]$
 which is obtained from $I$ by using  the $K$-algebra homomorphism $R\rightarrow R(\mathfrak{p})$  with $x_j\mapsto 1$
 for all $x_j\notin \{x_{i_1}, \ldots, x_{i_r}\}$.  It is well-known that
 $\mathfrak{p}\in \mathrm{Ass}_R(R/I)$ if and only if $\mathfrak{m}_\mathfrak{p}\in  \mathrm{Ass}(I(\mathfrak{p}))$,
 where $\mathfrak{m}_\mathfrak{p}$ is the graded maximal ideal of $R(\mathfrak{p})$. For  ease of reference, we recall the following theorem   that will be used to prove Theorem~\ref{NNTF-DI(CN)}.

 \begin{theorem} (\cite[Theorem 3.1]{NQBM})  \label{I+xH}
Let $I$ and $H$  be two  normal square-free monomial ideals in a  polynomial ring $R=K[x_1, \ldots, x_n]$
such that $I+H$ is normal.
  Let  $x_{c} \in \{x_1, \ldots, x_n\}$ be  a variable with
 $\gcd (v,x_c)=1$ for all $v\in \mathcal{G}(I)\cup  \mathcal{G}(H)$. Then $L:=I+x_cH$ is normal.
\end{theorem}

We are now  in the position to  state our next result related to the associated primes of powers of dominating ideals  of  cycle graphs.

 \begin{theorem}\label{NNTF-DI(CN)}
 Let $C_n$ be a cycle graph of order $n$ with  $V(C_n)=\{x_1, \ldots, x_n\}$ and $E(C_n)=\{\{x_i, x_{i+1}\}~:~ i=1, \ldots, n-1\} \cup \{\{x_n, x_1\}\}$.
 Let $DI(C_n)$ be the  dominating  ideal of $C_n$ in the polynomial ring $R=K[x_1, \ldots,x_n]$.   Then the following statements hold.
 \begin{itemize}
 \item[(i)] The monomial ideal $DI(C_n)$ satisfies the persistence property.
 \item[(ii)] For all $s\geq 1$, we have
 $$\mathrm{Ass}(DI(C_n)^s) \subseteq \{(x_i, x_{i+1}, x_{i+2})~:~i=1, \ldots, n\} \cup \{\mathfrak{m}\},$$
  where $\mathfrak{m}=(x_1, \ldots, x_n)$ and   $x_{n+1}$ (respectively, $x_{n+2}$) represents $x_1$ (respectively, $x_2$).
  \item[(iii)] $\mathrm{Ass}^{\infty }(DI(C_n))\subseteq \{(x_i, x_{i+1}, x_{i+2})~:~i=1, \ldots, n\}  \cup \{\mathfrak{m}\}.$
 \end{itemize}
  \end{theorem}

 \begin{proof}
For convenience of notation,  put $L:=DI(C_{n})$.  Since all claims are true for
$DI(C_3)=(x_1,x_2,x_3)$, we assume that $n\geq 4$.
We first show that   $\mathrm{Ass}(L^s) \subseteq \mathrm{Min}(L) \cup \{\mathfrak{m}\}$ for all $s\geq 1$. To  see this, fix $s\geq 1$, and pick an arbitrary element $\mathfrak{p}\in \mathrm{Ass}(L^s)$. If $\mathfrak{p}=\mathfrak{m}$, then there is nothing to show. We thus assume that $\mathfrak{p}\neq \mathfrak{m}$.
Without loss of generality, one may  assume that $\mathfrak{p}=(x_i~:~ i=1,  \ldots, r)$,  where $3\leq  r<n$.
Then we get  $L(\mathfrak{p})=\cap_{i=1}^{\ell}I^{\vee}_3(P_i)$, where $P_1, \ldots, P_\ell$ are some disjoint paths in $C_{n}$.   By virtue of
Lemma \ref{NTF-Alexander Dual-Path},   one can immediately conclude that  $I^{\vee}_3(P_i)$ is normally torsion-free for  all $i=1, \ldots, \ell$. In  the  light  of  Proposition \ref{Intersection}, we get  $L(\mathfrak{p})$ is normally torsion-free, and hence $\mathfrak{p}\in \mathrm{Min}(L(\mathfrak{p}))$.
On the other hand, based on \cite[Lemma 4.6]{RNA}(viii), we know that
$\mathfrak{p}\in \mathrm{Ass}(L^s)$  if and only if  $\mathfrak{p}\in \mathrm{Ass}(R(\mathfrak{p})/(L(\mathfrak{p})^s).$
Consequently, one obtains    $\mathfrak{p}\in \mathrm{Ass}(R(\mathfrak{p})/(L(\mathfrak{p})^s)$.
 We therefore get  $\mathfrak{p}=(x_{\alpha}, x_{\alpha+1}, x_{\alpha+2})$ for some $1\leq \alpha \leq r-2$, and thus $\mathfrak{p}\in \mathrm{Min}(L)$.   Accordingly, we derive that $\mathrm{Ass}(L^s) \subseteq \mathrm{Min}(L) \cup \{\mathfrak{m}\}$ for all $s\geq 1$.  This proves (ii).

 We now show that $L$ has the persistence property. Note that, in view of   the proof of  \cite[Corollary 4.10]{NQKR},   every normal monomial ideal has the  persistence property. Hence, it is sufficient for us to demonstrate  that $L$ is normal. Our strategy is to use Theorem \ref{I+xH}. To accomplish this,
 set $$A:=\bigcap_{i=1}^{n-3}(x_i, x_{i+1}, x_{i+2}) ~ \text{and} ~  B:=(x_{n-1}, x_{n-2})\cap (x_{n-1}, x_1) \cap (x_1, x_2).$$
      Then we have $L=A\cap (x_nR+ B)=x_nA+A\cap B$. Due to  $A$ is the Alexander dual of the $3$-path ideal  of   a  path  graph $P=(V(P), E(P))$  with the vertex set  $V(P)=\{x_1, \ldots, x_{n-1}\}$ and the following edge set $$E(P)=\{\{x_i, x_{i+1}\}~:~ i=1, \ldots, n-2\},$$   Lemma
      \ref{NTF-Alexander Dual-Path}  gives that $A$ is normally torsion-free, and by  \cite[Theorem 1.4.6]{HH1}, we deduce that $A$ is normal. Also, note that $A+A\cap B=A$ is normal. In what follows, our aim is to establish $A\cap B$ is normal.   It is easy  to check that
 \begin{align*}
 A\cap B& = \bigcap_{i=2}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2})\cap (x_{n-1}, x_1) \cap (x_1, x_2)  \\
  & =    x_1(\bigcap_{i=2}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2}))   \\
  &\quad     + x_2R \cap   \bigcap_{i=3}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2})\cap (x_{n-1}, x_1).
 \end{align*}
  To simplify the notation, put $C:=\bigcap_{i=2}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2})$ and
 $$D:=x_2R \cap   \bigcap_{i=3}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2})\cap (x_{n-1}, x_1).$$
 This gives rise to  $A\cap B=x_1C+D$.  As   the monomial ideal $\bigcap_{i=2}^{n-4}(x_i, x_{i+1}, x_{i+2})$ can be viewed as
 the Alexander dual of the  $3$-path ideal  of  a  path  graph,  by  Lemma \ref{NTF-Alexander Dual-Path}, is normally torsion-free, and on account of
 Lemma  \ref{NQBM}, we can conclude that $C$ is normally torsion-free, and based on   \cite[Theorem 1.4.6]{HH1}, is normal. Moreover, since $D \subseteq C$, one has $C+D=C$ is normal as well. Now, a similar argument shows that the monomial ideal $\bigcap_{i=3}^{n-4}(x_i, x_{i+1}, x_{i+2})$ is normally torsion-free, and after two times using  Lemma  \ref{NQBM}, we can derive that $\bigcap_{i=3}^{n-4}(x_i, x_{i+1}, x_{i+2})\cap (x_{n-1}, x_{n-2})\cap (x_{n-1}, x_1)$ is normally torsion-free. It follows now from
 \cite[Lemma 2.5]{ANKRQ} that $D$ is normal.  Thanks to  Theorem \ref{I+xH}, we get  $A\cap B= x_1C+D$ is normal. Due to $A$ and  $A\cap B$ are  normal, Theorem  \ref{I+xH} yields that  $L=x_nA+A\cap B$ is normal, and so has the persistence property, as we claimed in  (i).

 Finally, combining (i) and (ii) leads us to    $$\mathrm{Ass}^{\infty }(DI(C_n))\subseteq \{(x_i, x_{i+1}, x_{i+2})~:~i=1, \ldots, n\}  \cup \{\mathfrak{m}\}.$$ This demonstrates (iii), and the proof is complete.
 \end{proof}

Let $G$ be a simple finite graph. We finish this paper with a short argument on the relation between  prime monomial ideals  associated to  $\mathrm{Ass}(DI(G)^s)$ for some $s \geq 1$, and prime monomial ideals associated to  $\mathrm{Ass}(DI(G_\mathfrak{p})^s)$, where
 $G_\mathfrak{p}$ stands for the induced graph on $\mathfrak{p}$, i.e., the graph with vertex set  $\mathfrak{p}$, and edge set
 $E({G_{\mathfrak{p}}})=\{e\in E(G) \mid e\subseteq \mathfrak{p}\}$.

Francisco, Ha, and Van Tuyl, in \cite[Lemma 2.11]{FHV2}, showed that if $\mathcal{H}$ is a finite simple hypergraph on $V=\{x_1, \ldots, x_n\}$ with cover ideal $J(\mathcal{H})\subseteq R=k[x_1, \ldots, x_n]$, then for all $d\geq 1$,
  $$P=(x_{i_1}, \ldots, x_{i_r})\in \mathrm{Ass}(R/J(\mathcal{H})^d) \Leftrightarrow  P=(x_{i_1}, \ldots, x_{i_r})\in \mathrm{Ass}(k[P]/J(\mathcal{H}_P)^d),$$
  where $k[P]=k[x_{i_1}, \ldots, x_{i_r}]$, and  $\mathcal{H}_P$ is the induced hypergraph of $\mathcal{H}$ on the vertex set $P=\{x_{i_1}, \ldots, x_{i_r}\}\subseteq V$. Here,  we provide a counterexample which shows that one cannot mimic this result for dominating ideals of simple finite graphs. To do this, let $G=(V(G), E(G))$ be a simple finite graph with the vertex set $V(G)=\{x_1, x_2, x_3, x_4, x_5, x_6\}$ and the following edge set
\begin{align*}
  E(G)= & \{\{x_1, x_2\}, \{x_1, x_3\}, \{x_1, x_6\}, \{x_2, x_4\},\{x_2,x_5\}, \{x_3,x_4\}, \\
  &\{x_3,x_5\}, \{x_4,x_6\}\}.
  \end{align*}
  One can easily see that the dominating ideal of $G$ is given by
  \begin{align*}
  DI(G)= & (x_1,x_2,x_3,x_6) \cap (x_1,x_2,x_4,x_5)\cap (x_1,x_3,x_4,x_5)\cap (x_2,x_3,x_4,x_6)\\
  &\cap (x_2,x_3,x_5)\cap (x_1,x_4,x_6).
  \end{align*}
  Take  $\mathfrak{p}:=(x_1,x_2,x_3,x_6) \in \mathrm{Ass}(DI(G)^2)$. It follows from the definition that $V(G_\mathfrak{p})=\{x_1,x_2,x_3,x_6\}$ and
  $E(G_\mathfrak{p})=\{\{x_1, x_2\}, \{x_1, x_3\}, \{x_1, x_6\}\}$. In particular, the dominating ideal of $G_\mathfrak{p}$ is given by
  $$DI(G_\mathfrak{p})= (x_1,x_2) \cap (x_1,x_3)\cap (x_1,x_6)=(x_1, x_2x_3x_6).$$
  We can conclude from \cite[Theorem 2.5]{SN} that $DI(G_\mathfrak{p})$ is normally torsion-free, and so
  $\mathfrak{p} \notin \mathrm{Ass}(DI(G_\mathfrak{p})^2)$. On the other hand, $\mathfrak{q}:=(x_1,x_6)\in \mathrm{Ass}(DI(G_\mathfrak{p})^2)$, while $\mathfrak{q} \notin \mathrm{Ass}(DI(G)^2)$.

In what follows, we turn our attention to discuss the nature of  $G_{\mathfrak{p}}$. To accomplish this, we first recall the definition of  partial $t$-cover ideals which  has already been introduced  in \cite{BBV}.
\begin{definition} (\cite[Definition 1.1]{BBV})
Suppose that $G$ is a finite simple
graph on the vertex set $V(G) = \{x_1, x_2, \ldots, x_n\}$
 with the edge set $E(G)$. Also, for any $x\in V(G)$,  let
$N(x) = \{y~:~ \{x, y\}\in  E(G)\}$ denote the set of neighbors of $x$. Fix an integer $t\geq 1$. The {\it partial $t$-cover ideal} of $G$  is the monomial ideal
$$J_t(G)=\bigcap_{x\in V(G)}\left(\bigcap_{\{x_{i_1}, \ldots, x_{i_t}\} \subseteq N(x)}(x, x_{i_1}, \ldots, x_{i_t})\right).$$
\end{definition}

It should be noted that when $t=1$, the  construction above  is simply the cover ideal of a finite simple graph $G$.
In fact, Bhat, Biermann, and Van Tuyl, in  \cite[Lemma 2.4]{BBV}, proved the following lemma:

\begin{lemma} (\cite[Lemma 2.4]{BBV})
Let $G$ be a graph on the vertex set  $\{x_1, \ldots, x_n\}$, and $J_t(G)$ be the partial $t$-cover ideal  of $G$. If  $\mathfrak{p}=(x_{i_1}, \ldots, x_{i_r})\in \mathrm{Ass}(J_t(G)^s)$, then $G_\mathfrak{p}$ is connected.
\end{lemma}
In this direction,  we leave the following open question:
\begin{question}\label{OPEN-QUESTION}
 Let $G$ be a connected simple finite graph  on the vertex set $V(G)=\{x_1, \ldots, x_n\}$, and let $DI(G)$ be the  dominating  ideal of $G$. If
 $\mathfrak{p}=(x_{i_1}, \ldots, x_{i_r})\in \mathrm{Ass}(DI(G)^s)$ for some $s\geq 1$, then can we deduce that $G_\mathfrak{p}$ is connected?
\end{question}


Not everyone can drop a few hundred dollars on a VR headset, but that doesn't mean they can't experience VR! For those with the time and a bit of handiwork skill, this DIY guide from guest writer Ohaple will show you how to make a smartphone-based VR headset for as little as $20. Along the way, you'll learn the hardware and software basics of a VR headset.
Update (1/21/2016): An update to this guide provides new methods for lower latency gameplay using wired display mirroring software and a gyroscopic mouse for headtracking. Check it out here. Original guide continues below.
Virtual reality is on its way into the mainstream. As we see huge shifts such as the Oculus-Facebook buyout and Sony's Morpheus VR headset prototype, virtual reality is making its way into the hearts and minds of gamers everywhere. One problem still stands between the average consumer and a fun virtual reality experience: cost. Whether you follow a guide to create a DIY Oculus Rift or go out and buy the Oculus Rift DK2, you will find your wallet a few hundred dollars lighter. For those wanting a less expensive way to experience VR, using an existing smartphone for the basis of a VR headset greatly reduces the cost. In this guide, I'll show you how to make your own smartphone-based VR headset.
This DIY VR headset (also known as a head mounted display or HMD) takes a few hours to setup, and will cost somewhere between $20-$50 depending on how you source your parts and whether you want additional programs or controller support. If you already have a Durovis Dive, or would like to buy one, this project can be used with the Dive as well, just skip to the LED installation portion.
We will build a cheap and easy HMD, similar to the Durovis Dive.
We will use our computer to play stereoscopic 3D games. These can include Oculus Rift games, game mods (such as Minecrift), or games compatible with TriDef 3D, which takes 2D games and plays them in stereoscopic 3d.
We will then use our iOS or Android device as an external mirrored monitor to view the game in the VR headset.
Using an LED, we will establish head-tracking to control your computer's mouse, and consequently your character's head position in-game.
(1x) Pair of magnifying lenses. I used the frontmost lenses from a pair of binoculars ($3 at a thrift store) but if you are going to source them on the net, look for aspheric lenses with 5x magnification and about 1-2 inch diameter.
First, you need to determine how far your phone will need to sit from your face. Use your lenses to measure about how close the lenses give a good crisp focus. Mine took about 2 and 1/8 inches. This is the distance from your lenses to your display.
Measure the width of your reading glasses hinge to hinge onto the foam board. This will be the width of your HMD. Now that you have the focal length and the width, cut out 2 pieces of foam board as wide as the glasses, and 1.5x as long as the focal length. This is the top and bottom. of the unit. You can also make the bottom piece 1/4 inch longer than the top, to allow a shelf for the phone to sit on. If you don't. you can add it as a separate piece later.
Use a plate, or other round object to draw out a semi-circle on the top and bottom pieces. This is to allow it to fit nicely to your face. On the bottom piece, cut out an extra inch or two in the middle to allow venting under your nose (if you don't, the lenses will fog when you breathe through your nose).
Cut out 2 side pieces. They should both be as long as the top and bottom are (1.5x the focal length), and as tall as the front pieces are (the height of your phone in landscape mode. Mark the focal length on these pieces.
Cut out 2 front pieces. One of them will be solid, used to cover the back of the phone (to prevent false positive light sources from reflections on the gloss back of the phone). This is not needed if the back of your phone is non-reflective, or if you have a non-reflective phone case. The second one will have a hole cut in it the same size as the display of your phone, centered. These should be as wide as the top and bottom, and as tall as your phone is when in landscape mode.
You have all of the foam board cut out to size. Using hot glue, secure the top and bottom to the two sides. Then add the front piece with the cutout, and glue it in place. Make sure that you have left a shelf for the phone to sit on on the bottom piece, or add one now to the very bottom of the unit. This is to prevent your phone falling out the bottom. Glue the second front piece onto the edge of this shelf, making sure there is still room for your phone.
Pull the lenses out of the reading glasses. Pressure fit your new lenses into the frames, and glue in place.
Now for the lenses and elastic. Holes should be cut approximately 1/2 way up the side panels, at the focal length of your lenses.
Now cut the glasses at the hinge, and fit them through the two holes. Glue in place.
For the phone holder, glue a strip of elastic across the front of the unit. Be sure not to cover up where your phone charges, as you may need to plug in when you play long sessions. For the headstrap, secure a length of elastic to both sides using hot glue. Two strips may be required to hold it secure. You now have your completed HMD. Now to solve head-tracking. Put the phone in the elastic at the front, put it on, and see how it works. Go to the Durovis Dive website where you can find many apps intended for phone only so you can see how it works.
This is a wonderful project, and it's great that there is an exploration into wireless transmission video from a much stronger PC! If latencies can be reduced enough (encode/transmission/decode), or bandwidth increases enough, this may lead to very low cost headsets with great performance. What is most interesting is that cloud-rendering (eg. brigade3.com) suddenly seems much more realistic, and could blow the doors open for realism in VR at easy-to-explore prices.
Soon VR headsets will be as interchangeable as monitors or TVs. The hardware won't matter that much with some being better than others. What will matter is the system driving it and the eco-system of software.
Thanks for the nice comment, I agree with much of what you said. I am working right now to reduce latencies, and will have an update on my channel very soon to showcase it!
A suggestion for the LED for people who may not want or know how to solder. Get an LED reading light or keychain light from the dollar store. This way you can mount that to the top and still have an on/off switch.
free track doesn't seem to have a mac version. anything out there like it that works on mac?
Yes sir. The gyroscopic mouse. I have annotated the current video found in the writeup above to include a link to my newer video where I show how to incorporate a USB Gyroscopic mouse. This adds $10-20 to the project cost, but yields a much better result and is mac compatible.
What a cool project! I'm having some success right now with TriDef 3d launching Steam, and then using Android Limelight to use Nvidia's Shield stream to my Galaxy Note 3. The parts are on the way for the headset, I can't wait to try to build this, thanks for putting up the instructions man!
Sounds like you have the hang of it! I will be interested to see how it turns out for you. Check my channel for more updates to the project.
Just got all my parts in ( using the android gyro-mouse) and all seems to be working except tridef. Can't figure out what I'm doing wrong. Halo doesn't work, and team fortress 2 through steam isn't working. And it only seems to be on games, everything else in tridef works fine. I'm running windows 7 bootcamp on a 2013 Mac mini.
Some people have reported issues using splashtop with tridef 3d if you do not have a Geforce graphics card. Try using kainy in stead of splashtop if you are on android.
I am having trouble getting my mouse to move with the led. The program free track won't let me enable my mouse. Also I am having trouble with the splash program also. I really need help. I've built the box and it is awesome with dive games, but that's all I can play. I've made it this far. Help me make it all the way.
Great job! You are using aspheric lenses? what FOV do you get?
For everyone in the comments, it is much better to contact me on youtube or twitter, as I do not check this as often.
@Anibal I don't know how to calculate the FOV, but it seems similar to sitting 12 inches from my 21 inch monitor. I do not know if they are aspheric, as they were taken from binoculars. They do not distort the shapes.
Im going to try building this project, but i cant seem to find any 5x 1-2″ Lenses anywhere online. Could i just the lenses already in 5x reading glasses, or must i use round magnifier lenses?
Visit my blog: http://www.virtualystudio.blogspot.com , how make a hmd with a smartphone, i think that use freetrack is the past, 80% of smartphones have gyroscope. Using Freepie application you can move the cursor of windows with the movement of your smartphone.
A few years ago there was a guy called Johnny Lee who got excellent head tracking using a wii remote (check out here: http://youtu.be/Jd3-eiid-Uw?list=UUTW58h2zg8r9O0bGmMt2L5Q ) is there any way this method could be used for head tracking here as i'd say the results would be better than current method?
With Oculus rift games, I cant look up and down with the Freetrack mouse control.
you really want a current limiting resistor in your led circuit, adds minimal complexity and ensures desired brightness, maximum battery life, as well as greatly extends the life of the led. ideally you'd have some sort of dc-dc converter to ensure led brightness throughout the life of the battery, but if you aim for the voltage level at which the batteries are 50% drained or so you should be fine. msg me if you want further deets.
you don´t need one! a simple white LED has a normal working voltage of 3,4Volts at 20mA. since the led only gains 3 volt it will not even burn out or something like that. And to increase the lifetime isn't necessary, a basic LED will keep up 20.000 hours or way more. about the battery, a basic battery from Duracell or so will contain about 2500mA so they will be empty after 125 hours or so. thats long enough.
Actually you most certainly do need one. Unlike normal incandescent light bulbs, which are voltage-controlled (more voltage = more brightness), LEDs are current-controlled. They will draw as much current as they can regardless of the voltage applied, and if there's no current-limiting resistor, they'll draw as much as the battery can supply and burn out very quickly or even instantly.
tldr; you absolutely do need a current-limiting resistor.
your wrong, trust me, where did you get this information?
Wtf? It's common knowledge. The only two times you don't need a series current-limiting resistor with an LED is when the battery itself has sufficient resistance (such as in certain coin-cell battery-powered keychain lights), or if the LED has a built-in resistor (usually to save space on PCBs).
Try simply powering an LED from an unrestricted power source and tell me that it doesn't instantly burn out.
I had always used a resistor with LEDs, so I just tried an experiment, feeding 3v to a white LED with no resistor while measuring the current using a voltage & current controlled power supply. If the voltage was between 2.75v and 3.0v, the LED drew a limited amount of current (~15ma), but as the voltage increased above 3.1v, the current climbed dramatically.
At 3.3v it almost doubled to 35ma. At 3.5v it almost doubled again to 60ma.!
Conclusion, if your battery voltage stays below 3v, current isn't bad, but battery life will be significantly shorter than with a resistor.
Interesting! So it seems that you _can_ use an LED without a resistor, but it depends greatly on the particular LED and battery you have (i.e. the inherent resistance of the battery), and relies on very specific voltages.
In my previous tests of running LEDs without resistors (at 5v), some of them just flickered once and died, and others burned extremely brightly for a second or so then died. Never thought to take it further and try lower voltages.


Ready to meet the new Boyz on the block?
IMFACT is known for writing, choreographing, composing, and arranging many of their own tracks since their debut back in 2016.
We wanted to shine a light on some of the groups who will take center stage at the Staples Center next month. That includes NU'EST W, Pledis Entertainment's boy group that won over the hearts of many around the world with their dedication to music and each other.
o get attendees prepared for all of the amazing artists at KCON, we wanted to shine a light on some of the groups who will take center stage at the Staples Center next month. That includes Golden Child, Woollim Entertainment's newest boy group with members who are brighter than gold!
IN2IT, MMO & Stone Music Entertainment's newest boy group, had to overcome incredible odds to be put together.
In the mood to discover your new favorite ballad duo? Then look no further than monogram, a group that is a true hidden gem in the K-Indie music industry.
Which groups have had a detective concept? Here are a few of K-Pop's greatest groups of detectives!
Whether it's experimenting in new genres or continuing on their path of slow-grooving R&B and EDM songs, we know we can't wait to see what direction JJ Project goes with their next album.
Ready to step into the diamond life?
Apr 23 Party Like It's Your K-Pop Birthday!
Is there anything better than celebrating your birthday?
Everyone's favorite llama, Amber, is back with new music!
The story of UNB is one of determination, heartbreak, and renewal.
It's going to be a busy few months for CBX.
Released on March 23, "Super Duper" marks the 50th single from SM Entertainment's SM Station project.
Need another reason for why N.Flying should be on every K-Pop listener's radar?
Heaven is a place on Earth for NCTzens right now.
This past weekend, the group also began their "Busking Project," where they perform for three weeks at different locations in Seoul for fans.


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Q: H2G2: Why three pints? Was the 3 pints of beer Ford wanted Arthur to drink, for space travel preparation? Or the shock he may feel, when the Vogons arrive and destroy earth?
I assumed for a few years, it was for a more "comfortable" feeling of being teleported off earth into space, but now I am not sure.

A: Ford says that it's a 'muscle relaxant', presumably to help he and Arthur cope with the delitirious side-effects of teleportation.

Arthur: Why three pints?
Ford: Muscle relaxants. You'll need it.

A: In addition to Valorum's answer, when Arthur wakes up on the Vogon ship feeling terrible Ford remarks:

If you've never been through a matter transference beam before, you've probably lost some salt and protein. The beer you've had should have cushioned your system a bit.

redpajama_set_name: RedPajamaStackExchange


It's cold outside and time to bundle up!
Fashion expert Melissa Garcia shows us how to stay warm in these hot celebrity looks.
Tracksuits are back in a BIG way. Few have mastered how to wear a track suit, essentially sweats, with heels – few have mastered how to wear a track suit, essentially sweats, with heels, but Julianne Hough does it right.
The key to recreating this look was finding a track suit in the same color family and this one from NIKE does just that but at a fraction of the cost!. The navy coat draped over shoulders makes the look more sophisticated. This great little silver bag from ASOS for only $18 and a pair of navy heels pull the look together.
Winter White is another hot celebrity trend, worn recently by Rihanna.
Classy with Edge always looks chic. Kate Beckinsale's fur coat gives the entire look edge, but also a hefty price tag.
This look is all about the jacket and accessories.
Metallics are a huge trend this season and this look is all about this insane Acne Studio Gold Coat.
Recreate this look with a metallic bomber style coat and pair it with a faux fur vest underneath to create the effect of a metallic shearling coat!


The fast swimming keeps rolling with the Indianapolis Grand Prix, The Spanish Open, and Canadian Olympic Trials. We recap it all on SwimSwam News.
Garrett… This is the best report for someone like me. I'm busy (like most folks) with family, work, and everything else life throws at you, but I still want my swimming news. You deliver it fast, efficient, and in the most entertaining way. In my opinion, you are the best at this in the biz….. I've been a fan since your early days at floswimming. KEEP THEM COMING!


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Buy Law & Order: The Third Year DVD | GRUV
DVD Drama Television >
Law & Order: The Third Ye...ar [DVD]
Law & Order: The Third Year [DVD]
Your browser does Not support the video element.
Format Type: DVDCatalog No: 61120400UPC: 025192120671Rating: NRRelease Date: 06/03/2014Studio Description: Universal StudiosRegion: Region 1Running Time (mins): 1046Discs: 6 disc(s)Produced (year): 1993Category: DVD Drama Television
Powerful, timely and gripping, Law & Order has become television's most celebrated and widely acclaimed crime series of all time. Get ready for another thrilling season of ripped-from-the-headlines cases with all 22 episodes of the groundbreaking third season. Primetime Emmy® Award nominee Jerry Orbach joins New York's toughest team of prosecutors and investigators: Richard Brooks, Dann Florek, Steven Hill, Michael Moriarty, Chris Noth and Paul Sorvino. In the complex process of determining guilt and innocence, lives often hang in the balance: Law & Order: The Third Year is on the case.
Jerry Orbach, Steven Hill, Dann Florek, S. Epatha Merkerson, Paul Sorvino, Chris Noth, Michael Moriarty, Richard Brooks, Jill Hennessy
Covert Affairs: Season 5 [DVD]
Charlie's Angels - The Complete Series [DVD]
Comanche Moon [DVD]
Monk: Season 1 [DVD]
Alfred Hitchcock Presents: Season 2 [DVD]

redpajama_set_name: RedPajamaCommonCrawl


Uki is a wholesale fashion agency specialising in swimwear, located in south melbourne . Uki Fashion Agency Pty Ltd was founded in 1996 and covers Victoria, Tasmania, South Australia and the Northern Territory. Uki represents several leading swimwear collections including Baku ,Sea Level , Artesands , Genevieve, Bound , Bond-Eye , RAQ , Es Una and Sundaise accessories.


text:
namespace _02.ExtendedDatabase
{
    public class Startup
    {
        public static void Main()
        {
        }
    }
}

redpajama_set_name: RedPajamaGithub


text:
Protection of Employees (Fixed-Term Work) Act 2003
Written statements of employer.
8.—(1) Where an employee is employed on a fixed-term contract the fixed-term employee shall be informed in writing as soon as practicable by the employer of the objective condition determining the contract whether it is—
( a) arriving at a specific date,
( b) completing a specific task, or
( c) the occurrence of a specific event.
(2) Where an employer proposes to renew a fixed-term contract, the fixed-term employee shall be informed in writing by the employer of the objective grounds justifying the renewal of the fixed-term contract and the failure to offer a contract of indefinite duration, at the latest by the date of the renewal.
(3) A written statement under subsection (1) or (2) is admissible as evidence in any proceedings under this Act.
(4) If it appears to a rights commissioner or the Labour Court in any proceedings under this Act—
( a) that an employer omitted to provide a written statement, or
( b) that a written statement is evasive or equivocal,
the rights commissioner or the Labour Court may draw any inference he or she or it consider just and equitable in the circumstances.
Modifications (not altering text):
Application of section restricted (1.01.2018) by National Shared Services Office Act 2017 (26/2017), s. 18(8), S.I. No. 579 of 2017.
Membership and appointment of Board
(8) For the purposes of this Act only, members of the Board are not employees under section 8 and section 9 of the Protection of Employees (Fixed-Term Work) Act 2003 .


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Clonsilla is oorspronkelijk een plaats in het Ierse graafschap Dublin. Tegenwoordig is het een buitenwijk van de hoofdstad van het land en maakt bestuurlijk deel uit van Fingal. Clonsilla heeft een station aan de Dublin - Sligo dat bediend wordt door de forensentrein richting Maynooth. De M3, de snelweg van Dublin richting Navan loopt aan de noordkant langs Clonsilla.

Plaats in County Dublin

redpajama_set_name: RedPajamaWikipedia


BAC Beatbox Academy's 'Frankenstein: How to make a Monster' returns to the Battersea Arts Centre for a 3-week run, following their win of an Off-West End award last year. Billed as a retelling of Mary Shelley's classic this production is something of a misnomer: it draws on the themes of the well-known novel, with ideas about the dangers of technology and society's reaction to what is 'abnormal'. However, the associations are loose and it is clearly more a reaction to a shifting society. Elements of Frankenstein are a criticism of modern culture, calling out our obsession with phones and with social media as well as how easy it has become for the masses to pounce on those with bodies considered 'ugly'. At the same time, it is also a product of this very same culture: all performances are 'relaxed' and you are encouraged to take photos and videos throughout. Its hard to tell if this bitter irony is intentional.
The BAC's Grand Hall is the ideal venue for this production, with its sparse brick walls and the open emptiness of its stage. The set is simple, but incredibly effective, consisting only of bare lights hanging above the space occupied by the performers and large black speakers. The performers' outfits mirror the minimalism of the stage and all of them are dressed in simple grey hoodies and black jeans. The performers themselves, Aminita, Glitch, WIZ-RD, Native, ABH and Grove, are clearly massively talented, and its no surprise they rank amongst the UK's best in beatboxing. The real stand outs are ABH and Aminita. Aminita's unique singing voice adds a special quality to their performances that couldn't be achieved without her. ABH is arguably the most skilled member of the group, always drawing loud cheers from the audience with his solos.
Its hard to know what to make of this production. Its undeniably fun: even the bits of cheesy audience interaction scattered throughout the performance. However, the piece lacks any narrative structure and the ties to the themes of 'Frankenstein' are too loose to be tangible. If anything, this production is a showcase of BAC Beatbox Academy's talent, or of what can be achieved through beatbox as an artform. Frankenstein, though, it is not.


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Andy Murray tests positive for Covid-19, throwing Australian Open participation into doubt
Thursday January 14, 2021, 10:29 AM
Tennis star Andy Murray's participation in the upcoming Australian Open is in doubt after he tested positive for Covid-19.
The former world number one had been due to travel to Australia on one of the 18 charter flights laid on by tournament organisers but is stuck isolating at home.
ITV News understands that Murray - who is said to be in good health - is hoping to be able to arrive in Australia at a later date and participate in the year's first grand slam, which begins on February 8.
Andy Murray plays a shot against Switzerland's Stan Wawrinka in the first round match of the French Open in September. Credit: AP
Murray and his team are working closely with tournament director Craig Tiley to try to come up with an acceptable solution.
Organisers of the tournament have spent months negotiating arrangements with Australia's national government and local leaders in order to get more than 1,000 tennis players and associated personnel into the country.
Players are due to begin arriving in Australia within the next 24 hours before completing a two-week period of quarantine.
The plan is to allow players out of their rooms to practise for five hours a day.
Players had been told that a positive Covid test prior to flying would mean they were not allowed to travel to Australia.
Serbia's Novak Djokovic holds his trophy after winning the men's singles final at the Australian Open in 2020. Credit: AP
Last year's men's singles winner, Novak Djokovic, tested positive for Covid in June 2020 in the wake of the Adria Tour events he helped organise in Serbia and Croatia.
While other flagship events in the tennis calendar - including Wimbledon - fell victim to the disruption of the first wave of coronavirus.
The Australian Open 2020 hit the headlines after poor quality in the area impacted a number of players, following massive wildfires in the country.
Wimbledon cancelled 'due to public health concerns' amid coronavirus pandemic
Tennis player collapses on court at Australian Open due to 'poor air quality' from wildfires


On Saturday, Oct. 17 from 10 a.m. to 4 p.m., the Loudoun Campus of Northern Virginia Community College will host Loudoun County STEM Day. The event is free and open to the public.
Loudoun County STEM Day showcases opportunities in science, technology, engineering and math. Designed for all ages, the day features hands-on exhibits, workshops, food and music in a festival-like atmosphere.
STEM Day has educational workshops and exhibits for the whole family. Learn about robotics, game design, music recording, biotechnology, cybersecurity, engineering, horticulture, veterinary technology and more. Activities will be taught by NOVA and Loudoun County Public Schools faculty assisted by college and high school students.
The free event is a collaborative effort between Loudoun County Chamber of Commerce, Loudoun County Public Schools and NOVA.
NOVA-Loudoun is located at 21200 Campus Drive in Sterling (along Route 7). Follow signs to free parking. For more information, go to http://www.LoCoSTEM2015.com or call 703-948-7589.


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Zambia: Trial of activists for protesting government spending an affront to justice
24 September 2018, 02:05 UTC
The trial of six human rights defenders and activists - including well-known Zambian musician Pilato – simply for taking part in a peaceful protest against exorbitant government spending on fire trucks is politically-motivated and an affront to justice, said Amnesty International, as their hearing resumes in the Lusaka Magistrate's Court today.
These activists are facing trial simply for demanding transparency and accountability in public spending
Deprose Muchena, Amnesty International's Regional Director for Southern Africa
The six were charged with disobeying a lawful order after marching on parliament on 29 September 2017 against what they said was the corrupt procurement of 42 fire trucks for US$42 million. The reported cost of the trucks sparked public outcry over alleged misuse of public funds.
"These activists are facing trial simply for demanding transparency and accountability in public spending. They have committed no crime and should never have been charged in the first place," said Deprose Muchena, Amnesty International's Regional Director for Southern Africa.
Zambian authorities should respect, protect and promote the rights to freedom of expression and peaceful assembly in the country
Deprose Muchena
"The Zambian authorities must drop these politically motivated charges against them. The authorities should respect, protect and promote the rights to freedom of expression and peaceful assembly in the country."
The six, musician Fumba Chama who is known as Pilato, Lewis Mwape, Laura Miti, Sean Enock Tembo, Bornwell Mwewa and Mika Mwambazi have pleaded not guilty to the charges levelled against them.
If convicted, the human rights defenders and activists, who were beaten by arresting police officers, face up to two years in prison.


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text:
'Looks lame anyway': Elon Musk just deleted Facebook pages of Tesla, SpaceX
March 24, 2018 — 6.33am
To the delight of many of his Twitter followers, Tesla and SpaceX chief executive Elon Musk on Friday followed through on a promise to delete the Facebook pages of both companies, flushing more than 5 million combined "likes" down the digital drain on a whim.
The Facebook pages for the automaker and aerospace innovator have been replaced by a default page noting that "content isn't available right now."
"I didn't realise we had a Facebook page": Elon Musk
Musk apparently decided to take down the pages while responding to posts on Twitter, including one from Brian Acton, the co-founder of the Facebook-owned WhatsApp. Acton's tweet included the hashtag #deletefacebook.
That exchange led to a dare, as well as the revelation that Musk was unaware his aerospace company had a Facebook page tweeting:
"I didn't realise there was one. Will do."
It didn't take long for Musk to go on to delete Tesla's Facebook page, noting that the page "looks lame anyway."
By deleting (or ordering someone else to do so) his company's Facebook pages, Musk joined an online movement calling for Facebook users to sever their ties to the social media giant. The movement began with the revelation that Cambridge Analytica - a political marketing firm that worked with Donald Trump's presidential campaign - improperly harvested the data of 50 million Facebook users, raising new questions about the network's ability to protect user data.
SpaceX and Tesla have not disappeared from social media entirely. Accounts for both companies remain active on Instagram, which is owned by Facebook.
"Instagram's probably ok imo, so long as it stays fairly independent," Musk wrote on Twitter after calls for him to delete Tesla and SpaceX accounts on Instagram mounted Friday. "I don't use FB & never have, so don't think I'm some kind of martyr or my companies are taking a huge blow."
"Also," he added, "we don't advertise or pay for endorsements, so . . . don't care."
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